A complete solution to problems in an introduction to quantum field theory by peskin and schroeder

A Complete Solution to Problems in

“An Introduction to Quantum Field Theory”

by Peskin and Schroeder

Zhong-Zhi Xianyu

Harvard University

May 2016

ii

Preface

In this note I provide solutions to all problems and final projects in the book An Intro-

duction to Quantum Field Theory by M. E. Peskin and D. V. Schroeder [1], which I worked

out and typed into TEX during the first two years of my PhD study at Tsinghua University.

I once posted a draft version of them on my personal webpage using a server provided by

Tsinghua, which was however closed unfortunately after I graduated. Since then I received

quite a number of emails asking for the solutions, so I decided to put them on arXiv.∗

Nothing much has been updated in this note compared with the previous draft due to

the lack of time, except for some editorial work, as well as a few newly added references.

In particular, I don’t have enough time to proofread and therefore I cannot guarantee the

correctness of them, though I expect that most of them are correct. With that said, any

feedback via email† about errors, either physical or typographical, is much appreciated.

I would not claim any novelty or originality of this note, since almost all of problems in

the book belong to standard material of quantum field theory. Occasionally, I learned the

answer to a problem or the strategy for solving it before I started to work it out. But still,

I believe that the problem set in the book will always remain a treasure to any beginner of

this subject, and I feel it worthy to write up the solutions.

The contraction macro provided by the authors of the book‡ has been used in this note.

I would like to express my gratitude to Prof. Qing Wang and Prof. Hong-Jian He for

their wonderful courses of quantum field theory and their great help in my early days of

learning this subject. I would also like to thank Prof. Michael Peskin in particular, for his

generous permission and kind encouragement to letting me publish this note.

Comments on notations. All notations and conventions are the same with the book.

The book will be cited in the main text as “P&S” for short. The +iε prescription for

Feynman propagators is always assumed and is usually hidden.

∗The submission, however, was rejected by one of arXiv volunteer moderators based on the reason that

“arXiv does not allow submissions containing solutions to problems in physics textbooks”, and that “(the)

moderators consider that this type of submissions are harmful for students and instructors”. Insofar as I

can see, however, the solution can only do harm to those who are willing to do harm to themselves.†[email protected]‡http://physics.weber.edu/schroeder/qftbook.html

iii

iv Preface

Contents

Preface iii

2 The Klein-Gordon Field 1

2.1 Classical electromagnetism . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

2.2 The complex scalar field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

2.3 The spacelike correlation function . . . . . . . . . . . . . . . . . . . . . . . . 6

3 The Dirac Field 7

3.1 Lorentz group . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

3.2 The Gordon identity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

3.3 The spinor products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

3.4 Majorana fermions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

3.5 Supersymmetry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

3.6 Fierz transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

3.7 The discrete symmetries P , C and T . . . . . . . . . . . . . . . . . . . . . . 19

3.8 Bound states . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

4 Interacting Fields and Feynman Diagrams 23

4.1 Scalar field with a classical source . . . . . . . . . . . . . . . . . . . . . . . . 23

4.2 Decay of a scalar particle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

4.3 Linear sigma model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

4.4 Rutherford scattering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

5 Elementary Processes of Quantum Electrodynamics 33

5.1 Coulomb scattering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

5.2 Bhabha scattering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

5.3 The spinor products (2) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

5.4 Positronium lifetimes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

5.5 Physics of a massive vector boson . . . . . . . . . . . . . . . . . . . . . . . . 45

5.6 The spinor products (3) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

6 Radiative Corrections: Introduction 51

6.1 Rosenbluth formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

v

vi CONTENTS

6.2 Equivalent photon approximation . . . . . . . . . . . . . . . . . . . . . . . . 53

6.3 Exotic contributions to g − 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

7 Radiative Corrections: Some Formal Developments 59

7.1 Optical theorem in φ4 theory . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

7.2 Alternative regulators in QED . . . . . . . . . . . . . . . . . . . . . . . . . . 60

7.3 Radiative corrections in QED with Yukawa interaction . . . . . . . . . . . . 63

Final Project I. Radiation of Gluon Jets 67

9 Functional Methods 73

9.1 Scalar QED . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73

9.2 Statistical field theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74

10 Systematics of Renormalization 79

10.1 One-Loop structure of QED . . . . . . . . . . . . . . . . . . . . . . . . . . . 79

10.2 Renormalization of Yukawa theory . . . . . . . . . . . . . . . . . . . . . . . 80

10.3 Field-strength renormalization in φ4 theory . . . . . . . . . . . . . . . . . . . 82

10.4 Asymptotic behavior of diagrams in φ4 theory . . . . . . . . . . . . . . . . . 83

11 Renormalization and Symmetry 87

11.1 Spin-wave theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87

11.2 A zeroth-order natural relation . . . . . . . . . . . . . . . . . . . . . . . . . 88

11.3 The Gross-Neveu model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91

12 The Renormalization Group 93

12.1 Beta Function in Yukawa Theory . . . . . . . . . . . . . . . . . . . . . . . . 93

12.2 Beta Function of the Gross-Neveu Model . . . . . . . . . . . . . . . . . . . . 93

12.3 Asymptotic Symmetry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95

13 Critical Exponents and Scalar Field Theory 99

13.1 Correlation-to-scaling exponent . . . . . . . . . . . . . . . . . . . . . . . . . 99

13.2 The exponent η . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100

13.3 The CPN model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100

Final Project II. The Coleman-Weinberg Potential 103

15 Non-Abelian Gauge Invariance 113

15.1 Brute-force computations in SU(3) . . . . . . . . . . . . . . . . . . . . . . . 113

15.2 Adjoint representation of SU(2) . . . . . . . . . . . . . . . . . . . . . . . . . 113

15.3 Coulomb potential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114

15.4 Scalar propagator in a gauge theory . . . . . . . . . . . . . . . . . . . . . . . 115

15.5 Casimir operator computations . . . . . . . . . . . . . . . . . . . . . . . . . 117

CONTENTS vii

16 Quantization of Non-Abelian Gauge Theories 121

16.1 Arnowitt-Fickler gauge . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121

16.2 Scalar field with non-Abelian charge . . . . . . . . . . . . . . . . . . . . . . 122

16.3 Counterterm relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123

17 Quantum Chromodynamics 133

17.1 Two-Loop renormalization group relations . . . . . . . . . . . . . . . . . . . 133

17.2 A Direct test of the spin of the gluon . . . . . . . . . . . . . . . . . . . . . . 134

17.3 Quark-gluon and gluon-gluon scattering . . . . . . . . . . . . . . . . . . . . . 135

17.4 The gluon splitting function . . . . . . . . . . . . . . . . . . . . . . . . . . . 139

17.5 Photoproduction of heavy quarks . . . . . . . . . . . . . . . . . . . . . . . . 141

17.6 Behavior of parton distribution functions at small x . . . . . . . . . . . . . . 142

18 Operator Products and Effective Vertices 145

18.1 Matrix element for proton decay . . . . . . . . . . . . . . . . . . . . . . . . . 145

18.2 Parity-violating deep inelastic form factor . . . . . . . . . . . . . . . . . . . 147

18.3 Anomalous dimensions of gluon twist-2 operators . . . . . . . . . . . . . . . 151

18.4 Deep inelastic scattering from a photon . . . . . . . . . . . . . . . . . . . . . 156

19 Perturbation Theory Anomalies 159

19.1 Fermion number nonconservation in parallel E and B fields . . . . . . . . . . 159

19.2 Weak decay of the pion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161

19.3 Computation of anomaly coefficients . . . . . . . . . . . . . . . . . . . . . . 162

19.4 Large fermion mass limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164

20 Gauge Theories with Spontaneous Symmetry Breaking 167

20.1 Spontaneous breaking of SU(5) . . . . . . . . . . . . . . . . . . . . . . . . . 167

20.2 Decay modes of the W and Z bosons . . . . . . . . . . . . . . . . . . . . . . 168

20.3 e+e− →hadrons with photon-Z0 interference . . . . . . . . . . . . . . . . . . 170

20.4 Neutral-current deep inelastic scattering . . . . . . . . . . . . . . . . . . . . 173

20.5 A model with two Higgs fields . . . . . . . . . . . . . . . . . . . . . . . . . . 174

21 Quantization of Spontaneously Broken Gauge Theories 177

21.1 Weak-interaction contributions to the muon g − 2 . . . . . . . . . . . . . . . 177

21.2 Complete analysis of e+e− → W+W− . . . . . . . . . . . . . . . . . . . . . . 180

21.3 Cross section for du→ W−γ . . . . . . . . . . . . . . . . . . . . . . . . . . . 183

21.4 Dependence of radiative corrections on the Higgs boson mass . . . . . . . . . 185

Final Project III. Decays of the Higgs Boson 189

Bibliography 201

viii CONTENTS

Chapter 2

The Klein-Gordon Field

2.1 Classical electromagnetism

In this problem we derive the field equations and energy-momentum tensor from the

following action of classical electrodynamics,

S = − 1

4

∫d4xFµνF

µν , with Fµν = ∂µAν − ∂νAµ. (2.1)

(a) Maxwell’s equations To take variation of the classical action with respect to the

field Aµ, we note,

δFµνδ(∂λAκ)

= δλµδκν − δλν δκµ,

δFµνδAλ

= 0. (2.2)

Then from the first equality we get:

δ

δ(∂λAκ)

(FµνF

µν)

= 4F λκ. (2.3)

Now substitute this into Euler-Lagrange equation, we have,

0 = ∂µ

( δLδ(∂µAν)

)− δLδAν

= −∂µF µν . (2.4)

This is sometimes called the “second pair” of Maxwell’s equations. The so-called “first pair”

follows directly from the definition of Fµν = ∂µAν − ∂νAµ, and reads

∂λFµν + ∂µFνλ + ∂νFµλ = 0. (2.5)

The familiar electric and magnetic field strengths can be written as Ei = −F 0i and εijkBk =

−F ij, respectively. From this we deduce the Maxwell’s equations in terms of Ei and Bi:

∂iEi = 0, εijk∂jBk − ∂0Ei = 0, εijk∂jEk = 0, ∂iBi = 0. (2.6)

1

2 Chapter 2. The Klein-Gordon Field

(b) The energy-momentum tensor The energy-momentum tensor can be defined to be

the Nother current of the space-time translational symmetry. Under space-time translation

the vector Aµ transforms as,

δµAν = ∂µAν . (2.7)

Thus

T µν =∂L

∂(∂µAλ)∂νAλ − ηµνL = −F µλ∂νAλ +

1

4ηµνFλκF

λκ. (2.8)

Obviously, this tensor is not symmetric. We can add an additional term ∂λKλµν to T µν with

Kλµν antisymmetric with its first two indices. It’s easy to see that this term does not affect

the conservation of T µν . So if we choose Kλµν = F µλAν , then,

T µν = T µν + ∂λKλµν = F µλF ν

λ +1

4ηµνFλκF

λκ. (2.9)

Now this tensor is symmetric and is sometimes called the Belinfante tensor in literature.

We can also rewrite it in terms of Ei and Bi,

T 00 =1

2(EiEi +BiBi), T i0 = T 0i = εijkEjBk, etc. (2.10)

2.2 The complex scalar field

The Lagrangian is given by,

L = ∂µφ∗∂µφ−m2φ∗φ. (2.11)

(a) The conjugate momenta of φ and φ∗:

π =∂L∂φ

= φ∗, π =∂L∂φ∗

= φ = π∗. (2.12)

The canonical commutation relations:

[φ(x), π(y)] = [φ∗(x), π∗(y)] = iδ(x− y), (2.13)

The rest of commutators are all zero.

The Hamiltonian:

H =

∫d3x

(πφ+ π∗φ∗ − L

)=

∫d3x

(π∗π +∇φ∗ · ∇φ+m2φ∗φ

). (2.14)

(b) Now we Fourier transform the field φ as:

φ(x) =

∫d3p

(2π)3

1√2Ep

(ape

−ip·x + b†peip·x), (2.15)

thus:

φ∗(x) =

∫d3p

(2π)3

1√2Ep

(bpe−ip·x + a†pe

ip·x). (2.16)

2.2. The complex scalar field 3

Substitute the mode expansion into the Hamiltonian:

H =

∫d3x

(φ∗φ+∇φ∗ · ∇φ+m2φ∗φ

)=

∫d3x

∫d3p

(2π)3√

2Ep

d3q

(2π)3√

2Eq

×[EpEq

(a†pe

ip·x − bpe−ip·x)(aqe

−iq·x − b†qeiq·x)

+ p · q(a†pe

ip·x − bpe−ip·x)(aqe


+m2(a†pe

ip·x + bpe−ip·x

)(aqe

−iq·x + b†qeiq·x)]

=

∫d3x

∫d3p

(2π)3√

2Ep

d3q

(2π)3√

2Eq

×[(EpEq + p · q +m2)

(a†paqe

i(p−q)·x + bpb†qe−i(p−q)·x

)− (EpEq + p · q−m2)

(bqaqe

−i(p+q)·x + a†pb†qe

i(p+q)·x)]

=

∫d3p

(2π)3√

2Ep

d3q

(2π)3√

2Eq

×[(EpEq + p · q +m2)

(a†paqe

i(Ep−Eq)t + bpb†qe−i(Ep−Eq)t

)(2π)3δ(3)(p− q)

− (EpEq + p · q−m2)(bqaqe

−i(Ep+Eq)t + a†pb†qe

i(Ep+Eq)t)

(2π)3δ(3)(p + q)

]=

∫d3x

E2p + p2 +m2

2Ep

(a†pap + bpb

†p

)=

∫d3xEp

(a†pap + b†pbp + [bp, b

†p]), (2.17)

where we have used the mass-shell condition Ep =√m2 + p2. Note that the last term

contributes an infinite constant, which can be interpreted as the vacuum energy and can

be dropped, for instance, by the prescription of normal ordering. Then we get a finite

Hamiltonian,

H =

∫d3xEp

(a†pap + b†pbp

), (2.18)

Hence we get two sets of particles with the same mass m.

(c) The theory is invariant under the global transformation: φ→ eiθφ, φ∗ → e−iθφ∗. The

corresponding conserved charge is:

Q = i

∫d3x

(φ∗φ− φ∗φ

). (2.19)

Rewrite this in terms of the creation and annihilation operators:

Q = i

∫d3x

(φ∗φ− φ∗φ

)


= i

∫d3x

∫d3p

(2π)3√

2Ep

d3q

(2π)3√

2Eq

[(bpe−ip·x + a†pe

ip·x) ∂∂t

(aqe

−iq·x + b†qeiq·x)

− ∂

∂t


ip·x)·(aqe


=

∫d3x

∫d3p

(2π)3√

2Ep

d3q

(2π)3√

2Eq

[Eq


ip·x)(aqe


− Ep

(bpe−ip·x − a†peip·x

)(aqe


=

∫d3x

∫d3p

(2π)3√

2Ep

d3q

(2π)3√

2Eq

[(Eq − Ep)

(bpaqe

−i(p+q)·x − a†pb†qei(p+q)·x)

+ (Eq + Ep)(a†paqe

i(p−q)·x − bpb†qe−i(p−q)·x)]

=

∫d3p

(2π)3√

2Ep

d3q

(2π)3√

2Eq

×[(Eq − Ep)

(bpaqe

−i(Ep+Eq)t − a†pb†qei(Ep+Eqt))

(2π)3δ(3)(p + q)

+ (Eq + Ep)(a†paqe

i(Ep−Eq)t − bpb†qe−i(Ep−Eq)t)

(2π)3δ(3)(p− q)

]=

∫d3p

(2π)32Ep

· 2Ep(a†pap − bpb†p)

=

∫d3p

(2π)3

(a†pap − b†pbp

), (2.20)

where the last equal sign holds up to an infinitely large constant term, as we did when

calculating the Hamiltonian in (b). Then the commutators follow straightforwardly:

[Q, a†] = a†, [Q, b†] = −b†. (2.21)

We see that the particle a carries one unit of positive charge, and b carries one unit of

negative charge.

(d) Now we consider the case with two complex scalars of same mass. In this case the

Lagrangian is given by

L = ∂µΦ†i∂µΦi −m2Φ†iΦi, (2.22)

where Φi with i = 1, 2 is a two-component complex scalar. Then it is straightforward to

see that the Lagrangian is invariant under the U(2) transformation Φi → UijΦj with Uij a

matrix in fundamental representation of U(2) group. The U(2) group, locally isomorphic to

SU(2)×U(1), is generated by 4 independent generators 1 and 12τa, with τa Pauli matrices.

Then 4 independent Nother currents are associated, which are given by,

jµ =− ∂L∂(∂µΦi)

∆Φi −∂L

∂(∂µΦ∗i )∆Φ∗i = −(∂µΦ∗i )(iΦi)− (∂µΦi)(−iΦ∗i ),

2.2. The complex scalar field 5

jaµ =− ∂L∂(∂µΦi)

∆aΦi −∂L

∂(∂µΦ∗i )∆aΦ∗i = − i

2

[(∂µΦ∗i )τijΦj − (∂µΦi)τijΦ

∗j

]. (2.23)

The overall sign is chosen such that the particle carry positive charge, as will be seen in the

following. Then the corresponding Nother charges are given by,

Q =− i

∫d3x

(Φ∗iΦi − Φ∗i Φi

),

Qa =− i

2

∫d3x

[Φ∗i (τ

a)ijΦj − Φ∗i (τa)ijΦj

]. (2.24)

Repeating the derivations above, we can also rewrite these charges in terms of creation and

annihilation operators, as,

Q =

∫d3p

(2π)3

(a†ipaip − b

†ipbip

),

Qa =1

2

∫d3p

(2π)3

(a†ipτ

aijaip − b

†jpτ

aijbjp

). (2.25)

The generalization to N -component complex scalar is straightforward. In this case we

only need to replace the generators τa/2 of SU(2) group to the generators ta in the funda-

mental representation of SU(N) group with commutation relation [ta, tb] = ifabctc.

Then we are ready to calculate the commutators among all these Nother charges and

the Hamiltonian. Firstly we show that all charges of the U(N) group commute with the

Hamiltonian. For the U(1) generator, we have

[Q,H] =

∫d3p

(2π)3

d3q

(2π)3Eq

[(a†ipaip − b

†ipbip

),(a†jqajq + b†jqbjq

)]=

∫d3p

(2π)3

d3q

(2π)3Eq

(a†ip[aip, a

†jq]ajq + a†jq[a†ip, ajq]aip + (a→ b)

)=

∫d3p

(2π)3

d3q

(2π)3Eq

(a†ipaiq − a

†iqaip + (a→ b)

)(2π)3δ(3)(p− q)

= 0. (2.26)

Similar calculation gives [Qa, H] = 0. Then we consider the commutation among internal

U(N) charges:

[Qa, Qb] =

∫d3p

(2π)3

d3q

(2π)3

[(a†ipt

aijajp − b

†ipt

aijbjp

),(a†kqt

bkà`q − b

†kqt

bk`b`q

)]=

∫d3p

(2π)3

d3q

(2π)3

(a†ipt

aijt

bjà`q − a

†kqt

bk`t

a`jajp + (a→ b)

)(2π)3δ(3)(p− q)

= ifabc∫

d3p

(2π)3

(a†ipt

cijajp − b

†ipt

cijbjp

)= ifabcQc, (2.27)

and similarly, [Q,Q] = [Qa, Q] = 0.


2.3 The spacelike correlation function

We evaluate the correlation function of a scalar field at two points,

D(x− y) = 〈0|φ(x)φ(y)|0〉, (2.28)

with x− y being spacelike. Since any spacelike interval x− y can be transformed to a form

such that x0 − y0 = 0, thus we will simply take:

x0 − y0 = 0, and |x− y|2 = r2 > 0. (2.29)

Now:

D(x− y) =

∫d3p

(2π)3

1

2Epe−ip·(x−y) =

∫d3p

(2π)3

1

2√m2 + p2

eip·(x−y)

=1

(2π)3

∫ 2π

0

dϕ

∫ 1

−1

d cos θ

∫ ∞0

dpp2

2√m2 + p2

eipr cos θ

=−i

2(2π)2r

∫ ∞−∞

dppeipr√m2 + p2

(2.30)

Now we make the path deformation on p-complex plane, as is shown in Figure 2.3 of P&S.

Then the integral becomes,

D(x− y) =1

4π2r

∫ ∞m

dρρe−ρr√ρ2 −m2

=m

4π2rK1(mr). (2.31)

Chapter 3

The Dirac Field

3.1 Lorentz group

The generators of Lorentz group satisfy the following commutation relation,

[Jµν , Jρσ] = i(gνρJµσ − gµρJνσ − gnuσJµρ + gµσJνρ). (3.1)

(a) Let us redefine the generators as Li = 12εijkJ jk (All Latin indices denote spatial

components), with Li generate rotations, and Ki generate boosts. The commutators of

them can be derived straightforwardly to be,

[Li, Lj] = iεijkLk, [Ki, Kj] = −iεijkLk. (3.2)

If we further define J i± = 12

(Li ± iKi), then the commutators become,

[J i±, Jj±] = iεijkJk±, [J i+, J

j−] = 0. (3.3)

Thus we see that the algebra of the Lorentz group is a direct sum of two identical algebra

su(2).

(b) It follows that we can classify the finite dimensional representations of the Lorentz

group by a pair (j+, j−), where j± = 0, 1/2, 1, 3/2, 2, · · · are labels of irreducible representa-

tions of SU(2).

We study two specific cases.

1. (12, 0). Following the definition, we have J i+ represented by 1

2σi and J i− represented by

0. This implies

Li = (J i+ + J i−) = 12σi, Ki = −i(J i+ − J i−) = − i

2σi. (3.4)

Hence a field ψ under this representation transforms as:

ψ → e−iθiσi/2−ηiσi/2ψ. (3.5)

7

8 Chapter 3. The Dirac Field

2. (12, 0). In this case, J i+ → 0, J i− → 1

2σi. Then

Li = (J i+ + J i−) = 12σi, Ki = −i(J i+ − J i−) = i

2σi. (3.6)

Hence a field ψ under this representation transforms as:

ψ → e−iθiσi/2+ηiσi/2ψ. (3.7)

We see that a field under the representation (12, 0) and (0, 1

2) are precisely the left-handed

spinor ψL and right-handed spinor ψR, respectively.

(c) Let us consider the case of (12, 1

2). To put the field associated with this representation

into a familiar form, we note that a left-handed spinor can also be rewritten as row, which

transforms under the Lorentz transformation as:

ψTLσ2 → ψTLσ

2(1 + i

2θiσi + 1

2ηiσi

). (3.8)

Then the field under the representation (12, 1

2) can be written as a tensor with spinor indices:

ψRψTLσ

2 ≡ V µσµ =

(V 0 + V 3 V 1 − iV 2

V 1 + iV 2 V 0 − V 3

). (3.9)

In what follows we will prove that V µ is in fact a Lorentz vector.

A quantity V µ is called a Lorentz vector, if it satisfies the following transformation law:

V µ → ΛµνV

ν , (3.10)

where Λµν = δµν − i

2ωρσ(Jρσ)µν in its infinitesimal form. We further note that:

(Jρσ)µν = i(δρµδσν − δρνδσµ). (3.11)

and also, ωij = εijkθk, ω0i = −ωi0 = ηi, then the combination V µσµ = V iσi + V 0 transforms

according to

V iσi →(δij −

i

2ωmn(Jmn)ij

)V jσi +

(− i

2ω0n(J0n)i0 −

i

2ωn0(Jn0)0

i

)V 0σi

=(δij − i

2εmnkθ

k(−i)(δmi δnj − δmj δni )

)V jσi +

(− iηi(−i)(−δni )

)V 0σi

=V iσi − εijkV iθjσk + V 0ηiσi,

V 0 → V 0 +(− i

2ω0n(J0n)0i −

i

2ωn0(Jn0)0i

)V i

= V 0 +(− iηi(iδni )

)V i = V 0 + ηiV i.

In total, we have

V µσµ →(σi − εijkθjσk + ηi

)V i + (1 + ηiσi)V 0. (3.12)

3.2. The Gordon identity 9

If we can reach the same conclusion by treating the combination V µσµ a matrix transforming

under the representation (12, 1

2), then our original statement will be proved. In fact:

V µσµ →(

1− i

2θjσj +

1

2ηjσj

)V µσµ

(1 +

i

2θjσj +

1

2ηjσj

)=(σi +

i

2θj[σi, σj] +

1

2ηjσi, σj

)V i + (1 + ηiσi)V 0

=(σi − εijkθjσk + ηi

)V i + (1 + ηiσi)V 0, (3.13)

as expected. Hence we proved that V µ is a Lorentz vector.

3.2 The Gordon identity

In this problem we derive the Gordon identity,

u(p′)γµu(p) = u(p′)( p′µ + pµ

2m+

iσµν(p′ν − pν)2m

)u(p). (3.14)

Let us start from the right hand side,

u(p′)( p′µ + pµ

2m+

iσµν(p′ν − pν)2m

)u(p)

=1

2mu(p′)

((p′µ + pµ) + iσµν(p′ν − pν)

)u(p)

=1

2mu(p′)

(ηµν(p′ν + pν)−

1

2[γµ, γν ](p′ν − pν)

)u(p)

=1

2mu(p′)

( 1

2γµ, γν(p′ν + pν)−

1

2[γµ, γν ](p′ν − pν)

)u(p)

=1

2mu(p′)

(/p′γµ + γµ/p

)u(p) = u(p′)γµu(p),

where we have used the commutator and anti-commutators of gamma matrices, as well as

the Dirac equation.

3.3 The spinor products

In this problem, together with the Problems 5.3 and 5.6, we will develop a formalism

that can be used to calculating scattering amplitudes involving massless fermions or vector

particles. This method can profoundly simplify the calculations, especially in the calcula-

tions of QCD. Here we will derive the basic fact that the spinor products can be treated as

the square root of the inner product of lightlike Lorentz vectors. Then, in Problem 5.3 and

5.6, this relation will be put in use in calculating the amplitudes with external spinors and

external photons, respectively.

To begin with, let kµ0 and kµ1 be fixed four-vectors satisfying k20 = 0, k2

1 = −1 and

k0 · k1 = 0. With these two reference momenta, we define the following spinors:

1. Let uL0 be left-handed spinor with momentum k0;


2. Let uR0 = /k1uL0;

3. For any lightlike momentum p (p2 = 0), define:

uL(p) =1√

2p · k0/puR0, uR(p) =

1√2p · k/

puL0. (3.15)

(a) We show that /k0uR0 = 0 and /puL(p) = /puR(p) = 0 for any lightlike p. That is, uR0 is a

massless spinor with momentum k0, and uL(p), uR(p) are massless spinors with momentum

p. This is quite straightforward,

/k0uR0 = /k0/k1uL0 = (2gµν − γνγµ)k0µk1νuL0 = 2k0 · k1uL0 − /k1/k0uL0 = 0, (3.16)

and, by definition,

/puL(p) =1√

2p · k0/p/puR0 =

1√2p · k0

p2uR0 = 0. (3.17)

In the same way, we can show that /puR(p) = 0.

(b) Now we choose k0µ = (E, 0, 0,−E) and k1µ = (0, 1, 0, 0). Then in the Weyl represen-

tation, we have:

/k0uL0 = 0 ⇒

0 0 0 0

0 0 0 2E

2E 0 0 0

0 0 0 0

uL0 = 0. (3.18)

Thus uL0 can be chosen to be (0,√

2E, 0, 0)T , and:

uR0 = /k1uL0 =

0 0 0 1

0 0 1 0

0 −1 0 0

−1 0 0 0

uL0 =

0

0

−√

2E

0

. (3.19)

Let pµ = (p0, p1, p2, p3), then:

uL(p) =1√

2p · k0/puR0

=1√

2E(p0 + p3)

0 0 p0 + p3 p1 − ip2

0 0 p1 + ip2 p0 − p3

p0 − p3 −p1 + ip2 0 0

−p1 − ip2 p0 + p3 0 0

uR0

=1√

p0 + p3

−(p0 + p3)

−(p1 + ip2)

0

0

. (3.20)

3.4. Majorana fermions 11

In the same way, we get:

uR(p) =1√

p0 + p3

0

0

−p1 + ip2

p0 + p3

. (3.21)

(c) We construct explicitly the spinor product s(p, q) and t(p, q).

s(p, q) = uR(p)uL(q) =(p1 + ip2)(q0 + q3)− (q1 + iq2)(p0 + p3)√

(p0 + p3)(q0 + q3); (3.22)

t(p, q) = uL(p)uR(q) =(q1 − iq2)(p0 + p3)− (p1 − ip2)(q0 + q3)√

(p0 + p3)(q0 + q3). (3.23)

It can be easily seen that s(p, q) = −s(q, p) and t(p, q) = (s(q, p))∗.

Now we calculate the quantity |s(p, q)|2:

|s(p, q)|2 =

(p1(q0 + q3)− q1(p0 + p3)

)2+(p2(q0 + q3)− q2(p0 + p3)

)2

(p0 + p3)(q0 + q3)

=(p21 + p2

2)q0 + q3

p0 + p3

+ (q21 + q2

2)p0 + p3

q0 + q3

− 2(p1q1 + p2q2)

=2(p0q0 − p1q1 − p2q2 − p3q3) = 2p · q. (3.24)

Where we have used the lightlike properties p2 = q2 = 0. Thus we see that the spinor

product can be regarded as the square root of the 4-vector dot product for lightlike vectors.

3.4 Majorana fermions

(a) We at first study a two-component massive spinor χ lying in (12, 0) representation,

transforming according to χ→ UL(Λ)χ. It satisfies the following equation of motion:

iσµ∂µχ− imσ2χ∗ = 0. (3.25)

To show this equation is indeed an admissible equation, we need to justify: 1) It is relativis-

tically covariant; 2) It is consistent with the mass-shell condition (namely the Klein-Gordon

equation).

To show the condition 1) is satisfied, we note that γµ is invariant under the simultaneous

transformations of its Lorentz indices and spinor indices. That is ΛµνU(Λ)γνU(Λ−1) = γµ.

This implies

ΛµνUR(Λ)σνUL(Λ−1) = σµ,

as can be easily seen in chiral basis. Then, the combination σµ∂µ transforms as σµ∂µ →UR(Λ)σµ∂µUL(Λ−1). As a result, the first term of the equation of motion transforms as

iσµ∂µχ→ iUR(Λ)σµ∂µUL(Λ−1)UL(Λ)χ = UR(Λ)[iσµ∂µχ

]. (3.26)


To show the full equation of motion is covariant, we also need to show that the second term

iσ2χ∗ transforms in the same way. To see this, we note that in the infinitesimal form,

UL = 1− iθiσi/2− ηiσi/2, UR = 1− iθiσi/2 + ηiσi/2.

Then, under an infinitesimal Lorentz transformation, χ transforms as:

χ→ (1− iθiσi/2− ηiσi/2)χ, ⇒ χ∗ → (1 + iθiσi/2− ηiσi/2)χ∗

⇒ σ2χ∗ → σ2(1 + iθi(σ∗)i/2− ηi(σ∗)i/2)χ∗ = (1− iθiσi/2 + ηiσi/2)σ2χ∗.

That is to say, σ2χ∗ is a right-handed spinor that transforms as σ2χ∗ → UR(Λ)σ2χ∗. Thus

we see the the two terms in the equation of motion transform in the same way under the

Lorentz transformation. In other words, this equation is Lorentz covariant.

To show the condition 2) also holds, we take the complex conjugation of the equation:

−i(σ∗)µ∂µχ∗ − imσ2χ = 0.

Combining this and the original equation to eliminate χ∗, we get

(∂2 +m2)χ = 0, (3.27)

which has the same form with the Klein-Gordon equation.

(b) Now we show that the equation of motion above for the spinor χ can be derived from

the following action through the variation principle:

S =

∫d4x

[χ†iσ · ∂χ+

im

2(χTσ2χ− χ†σ2χ∗)

]. (3.28)

Firstly, let us check that this action is real, namely S∗ = S. In fact,

S∗ =

∫d4x

[(χ†iσ · ∂χ)† − im

2(χ†σ2χ∗ − χTσ2χ)

],

where the first term (χ†iσ · ∂χ)† = −i(∂χ†)iσχ is identical to the original kinetic term upon

integration by parts. Thus we see that S∗ = S.

Now we vary the action with respect to χ†, that gives

0 =δS

δχ†= iσ · ∂χ− im

2· 2σ2χ∗ = 0, (3.29)

which is exactly the Majorana equation.

(c) Let us rewrite the Dirac Lagrangian in terms of two-component spinors:

L = ψ(i/∂ −m)ψ

=(χ†1 −iχT2 σ

2)(0 1

1 0

)(−m iσµ∂µ

iσµ∂µ −m

)(χ1

iσ2χ∗2

)= iχ†1σ

µ∂µχ1 + iχT2 σµ∗∂µχ

∗2 − im

(χT2 σ

2χ1 − χ†1σ2χ∗2)

= iχ†1σµ∂µχ1 + iχ†2σ

µ∂µχ2 − im(χT2 σ

2χ1 − χ†1σ2χ∗2), (3.30)

where the equality should be understood to hold up to a total derivative term.

3.4. Majorana fermions 13

(d) The familiar global U(1) symmetry of the Dirac Lagrangian ψ → eiαψ now becomes

χ1 → eiαχ1, χ2 → e−iαχ2. The associated Nother current is

Jµ = ψγµψ = χ†1σµχ1 − χ†2σµχ2. (3.31)

To show its divergence ∂µJµ vanishes, we make use of the equations of motion:

iσµ∂µχ1 − imσ2χ∗2 = 0,

iσµ∂µχ2 − imσ2χ∗1 = 0,

i(∂µχ†1)σµ − imχT2 σ

2 = 0,

i(∂µχ†2)σµ − imχT1 σ

2 = 0.

Then we have

∂µJµ = (∂µχ

†1)σµχ1 + χ†1σ

µ∂µχ1 − (∂µχ†2)σµχ2 − χ†2σµ∂µχ2

= m(χT2 σ

2χ1 + χ†1σ2χ∗2 − χT1 σ2χ2 − χ†2σ2χ∗1

)= 0. (3.32)

In a similar way, one can also show that the Nother currents associated with the global

symmetries of Majorana fields have vanishing divergence.

(e) To quantize the Majorana theory, we introduce the canonical anticommutation rela-

tion, χa(x), χ†b(y)

= δabδ

(3)(x− y),

and also expand the Majorana field χ into modes. To motivate the mode expansion, we

note that the Majorana Langrangian can be obtained by replacing the spinor χ2 in the

Dirac Lagrangian (3.30) with χ1. Then, according to our experience in Dirac theory, it can

be found that

χ(x) =

∫d3p

(2π)3

√p · σ2Ep

∑a

[ξaaa(p)e−ip·x + (−iσ2)ξ∗aa

†a(p)eip·x

]. (3.33)

Then with the canonical anticommutation relation above, we can find the anticommutators

between annihilation and creation operators:

aa(p), a†b(q) = δabδ(3)(p− q), aa(p), ab(q) = a†a(p), a†b(q) = 0. (3.34)

On the other hand, the Hamiltonian of the theory can be obtained by Legendre transforming

the Lagrangian:

H =

∫d3x

(δLδχ

χ− L)

=

∫d3x

[iχ†σ · ∇χ+

im

2

(χ†σ2χ∗ − χTσ2χ

)]. (3.35)

Then we can also represent the Hamiltonian H in terms of modes:

H =

∫d3x

∫d3pd3q

(2π)6√

2Ep2Eq

∑a,b

[(ξ†aa†a(p)e−ip·x + ξTa (iσ2)aa(p)eip·x

)


× (√p · σ)†(−q · σ)

√q · σ

(ξbab(q)eiq·x − (−iσ2)ξ∗ba

†b(q)e−iq·x

)+

im

2

(ξ†aa†a(p)e−ip·x + ξTa (iσ2)aa(p)eip·x

)× (√p · σ)†σ2(

√q · σ)∗

(ξ∗ba†b(q)e−iq·x + (−iσ2)ξbab(q)eiq·x

)− im

2

(ξTa aa(p)eip·x + ξ†a(iσ

2)a†a(p)e−ip·x)

× (√p · σ)Tσ2√q · σ

(ξbab(q)eiq·x + (−iσ2)ξ∗ba

†b(q)e−iq·x

)]=

∫d3x

∫d3pd3q

(2π)6√

2Ep2Eq

∑a,b

a†a(p)ab(q)ξ†a

[(√p · σ)†(−q · σ)

√q · σ

+im

2(√p · σ)†σ2(

√q · σ)∗(−iσ2)− im

2(iσ2)(

√p · σ)Tσ2√q · σ

]ξbe−i(p−q)·x

+ a†a(p)a†b(q)ξ†a

[− (√p · σ)†(−q · σ)

√q · σ(−iσ2) +

im

2(√p · σ)†σ2(

√q · σ)∗

− im

2(iσ2)(

√p · σ)Tσ2√q · σ(−iσ2)

]ξ∗b e−i(p+q)·x

+ aa(p)ab(q)ξTa

[(iσ2)(

√p · σ)†(−q · σ)

√q · σ +

im

2(iσ2)(

√p · σ)†σ2(

√q · σ)∗(−iσ2)

− im

2(√p · σ)Tσ2√q · σ

]ξbe

i(p+q)·x

+ aa(p)a†b(q)ξTa

[− (iσ2)(

√p · σ)†(−q · σ)

√q · σ(−iσ2) +

im

2(iσ2)(

√p · σ)†σ2(

√q · σ)∗

− im

2(√p · σ)Tσ2√q · σ(−iσ2)

]ξ∗b e

i(p−q)·x

=

∫d3p

(2π)32Ep

∑a,b

a†a(p)ab(p)ξ†a

[(√p · σ)†(−p · σ)

√p · σ

+im

2(√p · σ)†σ2(

√p · σ)∗(−iσ2)− im

2(iσ2)(

√p · σ)Tσ2√p · σ

]ξb

+ a†a(p)a†b(−p)ξ†a

[− (√p · σ)†(p · σ)

√p · σ(−iσ2) +

im

2(√p · σ)†σ2(

√p · σ)∗

− im

2(iσ2)(

√p · σ)Tσ2

√p · σ(−iσ2)

]ξ∗b

+ aa(p)ab(−p)ξTa

[(iσ2)(

√p · σ)†(p · σ)

√p · σ +

im

2(iσ2)(

√p · σ)†σ2(

√p · σ)∗(−iσ2)

− im

2(√p · σ)Tσ2

√p · σ

]ξb

+ aa(p)a†b(p)ξTa

[− (iσ2)(

√p · σ)†(−p · σ)

√p · σ(−iσ2) +

im

2(iσ2)(

√p · σ)†σ2(

√p · σ)∗

− im

2(√p · σ)Tσ2√p · σ(−iσ2)

]ξ∗b

3.5. Supersymmetry 15

=

∫d3p

(2π)32Ep

∑a,b

1

2

(E2p + |p|2 +m2

)[a†a(p)ab(p)ξ†aξb − aa(p)a†b(p)ξTa ξ

∗b

]=

∫d3p

(2π)3

Ep2

∑a

[a†a(p)aa(p)− aa(p)a†a(p)

]=

∫d3p

(2π)3Ep∑a

a†a(p)aa(p). (3.36)

In the calculation above, each step goes as follows in turn: (1) Substituting the mode ex-

pansion for χ into the Hamiltonian. (2) Collecting the terms into four groups, characterized

by a†a, a†a†, aa and aa†. (3) Integrating over d3x to produce a delta function, with which

one can further finish the integration over d3q. (4) Using the following relations to simplify

the spinor matrices:

(p · σ)2 = (p · σ)2 = E2p + |p|2, (p · σ)(p · σ) = p2 = m2, p · σ = 1

2(p · σ − p · σ).

In this step, the a†a† and aa terms vanish, while the aa† and a†a terms remain. (5)

Using the normalization ξ†aξb = δab to eliminate spinors. (6) Using the anticommutator

aa(p), a†a(p) = δ(3)(0) to further simplify the expression. In this step we have throw away

a constant term − 12Epδ

(3)(0) in the integrand. The minus sign of this term indicates that

the vacuum energy contributed by Majorana field is negative. With these steps done, we

find the desired result, as shown above.

3.5 Supersymmetry

(a) In this problem we briefly study the Wess-Zumino model, which may be the simplest

supersymmetric field theory in 4 dimensional spacetime. Firstly let us consider the massless

case, in which the Lagrangian is given by

L = ∂µφ∗∂µφ+ χ†iσµ∂µχ+ F ∗F, (3.37)

where φ is a complex scalar field, χ is a Weyl fermion, and F is a complex auxiliary scalar

field. By auxiliary we mean a field with no kinetic term in the Lagrangian and thus it does

not propagate, or equivalently, it has no particle excitation. However, in the following, we

will see that it is crucial to maintain the off-shell supersymmtry of the theory.

The supersymmetry transformation in its infinitesimal form is given by:

δφ = −iεTσ2χ, (3.38a)

δχ = εF + σµ(∂µφ)σ2ε∗, (3.38b)

δF = −iε†σµ∂µχ, (3.38c)

where ε is a 2-component Grassmann variable. Now let us show that the Lagrangian is

invariant (up to a total divergence) under this supersymmetric transformation. This can be

checked term by term, as follows:

δ(∂µφ∗∂µφ) = i

(∂µχ

†σ2ε∗)∂µφ+ (∂µφ

∗)(− iεTσ2∂µχ

),


δ(χ†iσµ∂µχ) =(F ∗ε† + εTσ2σν∂νφ

∗)iσµ∂µχ+ χ†iσµ(ε∂µF + σνσ2ε∗∂µ∂νφ

)= iF ∗ε†σ2∂µχ+ i∂µ

[εTσ2σν σµ(∂νφ

∗)χ]− iεTσ2σν σµ(∂ν∂µφ

∗)χ

+ iχ†σµε∂µF + iχ†σµσνσ2ε∗∂µ∂νφ

= iF ∗ε†σ2∂µχ+ i∂µ[εTσ2σν σµ(∂νφ

∗)χ]− iεTσ2(∂2φ∗)χ

+ iχ†σµε∂µF + iχ†σ2ε∗∂2φ,

δ(F ∗F ) = i(∂µχ†)σµεF − iF ∗ε†σµ∂µχ,

where we have used σµσν∂µ∂ν = ∂2. Now summing the three terms above, we get:

δL = i∂µ

[χ†σ2ε∗∂µφ+ χ†σµεF + φ∗εTσ2

(σµσνσν − ∂µ

)χ], (3.39)

which is indeed a total derivative.

(b) Now let us add the mass term in to the original massless Lagrangian:

∆L =(mφF + 1

2imχTσ2χ

)+ c.c. (3.40)

Let us show that this mass term is also invariant under the supersymmetry transformation,

up to a total derivative:

δ(∆L) =− imεTσ2χF − imφε†σµ∂µχ+ 12

im[εTF + ε†(σ2)T (σµ)T∂µφ]σ2χ

+ 12

imχTσ2[εF + σµ(∂µφ)σ2ε∗] + c.c.

=− 12imF (εTσ2χ− χTσ2ε)− imφε†σµ∂µχ

− 12

im(∂µφ)ε†σµχ+ 12

im(∂µφ)χT (σµ)T ε∗ + c.c.

=− 12imF (εTσ2χ− χTσ2ε)− im∂µ(φε†σµχ)

+ 12

im(∂µφ)[ε†σµχ+ χT (σµ)T ε∗] + c.c

=− im∂µ(φε†σµχ) + c.c (3.41)

where we have used the following relations:

(σ2)T = −σ2, σ2(σµ)Tσ2 = σµ, εTσ2χ = χTσ2ε, ε†σµχ = −χT (σµ)T ε∗.

Now let us write down the Lagrangian with the mass term:

L = ∂µφ∗∂µφ+ χ†iσµ∂µχ+ F ∗F +

(mφF + 1

2imχTσ2χ+ c.c.

). (3.42)

Varying the Lagrangian with respect to F ∗, we get the corresponding equation of motion:

F = −mφ∗. (3.43)

Substitute this algebraic equation back into the Lagrangian to eliminate the field F , we get

L = ∂µφ∗∂µφ−m2φ∗φ+ χ†iσµ∂µχ+ 1

2

(imχTσ2χ+ c.c.

). (3.44)

Thus we see that the scalar field φ and the spinor field χ have the same mass.

3.5. Supersymmetry 17

(c) We can also include interactions into this model. Generally, we can write a Lagrangian

with nontrivial interactions containing fields φi, χi and Fi (i = 1, · · · , n), as

L = ∂µφ∗i∂

µφi + χ†i iσµ∂µχi + F ∗i Fi +

[Fi∂W [φ]

∂φi+

i

2

∂2W [φ]

∂φi∂φjχTi σ

2χj + c.c.

], (3.45)

where W [φ] is an arbitrary function of φi.

To see this Lagrangian is supersymmetry invariant, we only need to check the interactions

terms in the square bracket:

δ

[Fi∂W [φ]

∂φi+

i

2

∂2W [φ]

∂φi∂φjχTi σ

2χj + c.c.

]=− iε†σµ(∂µχi)

∂W

∂φi+ Fi

∂2W

∂φi∂φj(−iεTσ2χj) +

i

2

∂3W

∂φi∂φj∂φk(−iεTσ2χk)χ

Ti σ

2χj

+i

2

∂2W

∂φi∂φj

[(εTFi + ε†(σ2)T (σµ)T∂µφi

)σ2χj + χTi σ

2(εFj + σµ∂µφjσ

2ε∗)]

+ c.c..

The term proportional to ∂3W/∂φ3 vanishes. To see this, we note that the partial derivatives

with respect to φi are commutable, hence ∂3W/∂φi∂φj∂φk is totally symmetric on i, j, k.

However, we also have the following identity:

(εTσ2χk)(χTi σ

2χj) + (εTσ2χi)(χTj σ

2χk) + (εTσ2χj)(χTk σ

2χi) = 0, (3.46)

which can be directly checked by brute force. Then it can be easily seen that the ∂3W/∂φ3

term vanishes indeed. On the other hand, the terms containing F also sum to zero, which

is also straightforward to justify. Hence the terms left now are

− iε†σµ(∂µχi)∂W

∂φi+ i

∂2W

∂φi∂φjε†(σ2)T (σµ)T (∂µφi)σ

2χj

=− i∂µ

(ε†σµχi

∂W

∂φi

)+ iε†σµχi

∂2W

∂φi∂φj∂µφj − i

∂2W

∂φi∂φjε†σµ(∂µφi)χj

=− i∂µ

(ε†σµχi

∂W

∂φi

), (3.47)

which is a total derivative. Thus we conclude that the Lagrangian (3.45) is supersymmetri-

cally invariant up to a total derivative.

Let us end up with a explicit example, in which we choose n = 1 and W [φ] = gφ3/3.

Then the Lagrangian (3.45) becomes

L = ∂µφ∗∂µφ+ χ†iσµ∂µχ+ F ∗F +

(gFφ2 + iφχTσ2χ+ c.c.

). (3.48)

We can eliminate F by solving it from its field equation,

F + g(φ∗)2 = 0. (3.49)

Substituting this back into the Lagrangian, we get

L = ∂µφ∗∂µφ+ χ†iσµ∂µχ− g2(φ∗φ)2 + ig(φχTσ2χ− φ∗χ†σ2χ∗). (3.50)

This is a Lagrangian of massless complex scalar and a Weyl spinor, with φ4 and Yukawa

interactions. The field equations can be easily got by the variation.


3.6 Fierz transformations

In this problem, we derive the generalized Fierz transformation, with which one can

express (u1ΓAu2)(u3ΓBu4) as a linear combination of (u1ΓCu4)(u3ΓDu2), where ΓA is any

normalized Dirac matrices from the following set:1, γµ, σµν = i

2[γµ, γν ], γ5γµ, γ5 = −iγ0γ1γ2γ3

.

(a) The Dirac matrices ΓA are normalized according to

tr (ΓAΓB) = 4δAB. (3.51)

For instance, the unit element 1 is already normalized, since tr (1·1) = 4. For Dirac matrices

containing one γµ, we calculate the trace in Weyl representation without loss of generality.

Then the representation of

γµ =

(0 σµ

σµ 0

)gives tr (γµγµ) = 2 tr (σµσµ) (no sum on µ). For µ = 0, we have tr (γ0γ0) = 2 tr (12×2) = 4,

and for µ = i = 1, 2, 3, we have tr (γiγi) = −2 tr (σiσi) = −2 tr (12×2) = −4 (no sum on i).

Thus the normalized gamma matrices are γ0 and iγi.

In the same way, we can work out the rest of the normalized Dirac matrices, as:

tr (σ0iσ0i) = −2 tr (σiσi) = −4, (no sum on i)

tr (σijσij) = 2 tr (σkσk) = 4, (no sum on i, j, k)

tr (γ5γ5) = 4,

tr (γ5γ0γ5γ0) = −4, tr (γ5γiγ5γi) = 4.

Thus the 16 normalized elements are:1, γ0, iγi, iσ0i, σij, γ5, iγ5γ0, γ5γi

. (3.52)

(b) Now we derive the desired Fierz identity, which can be written as:

(u1ΓAu2)(u3ΓBu4) =∑C,D

CABCD(u1ΓCu4)(u3ΓDu2). (3.53)

Left-multiplying the equality by (u2ΓFu3)(u4ΓEu1), we get:

(u2ΓFu3)(u4ΓEu1)(u1ΓAu2)(u3ΓBu4) =∑CD

CABCD tr (ΓEΓC) tr (ΓFΓD). (3.54)

The left hand side:

(u2ΓFu3)(u4ΓEu1)(u1ΓAu2)(u3ΓBu4) = u4ΓEΓAΓFΓBu4 = tr (ΓEΓAΓFΓB);

the right hand side:∑C,D

CABCD tr (ΓEΓC) tr (ΓFΓD) =

∑C,D

CABCD4δEC4δFD = 16CAB

EF ,

thus we conclude:

CABCD = 1

16tr (ΓCΓAΓDΓB). (3.55)

3.7. The discrete symmetries P , C and T 19

(c) Now we derive two Fierz identities as particular cases of the results above. The first

one is:

(u1u2)(u3u4) =∑C,D

tr (ΓCΓD)

16(u1ΓCu4)(u3ΓDu2). (3.56)

The traces on the right hand side do not vanish only when C = D, thus we get:

(u1u2)(u3u4) =∑C

14

(u1ΓCu4)(u3ΓCu2)

= 14

[(u1u4)(u3u2) + (u1γ

µu4)(u3γµu2) + 12

(u1σµνu4)(u3σµνu2)

− (u1γ5γµu4)(u3γ

5γµu2) + (u1γ5u4)(u3γ

5u2)]. (3.57)

The second example is:

(u1γµu2)(u3γµu4) =

∑C,D

tr (ΓCγµΓDγµ)

16(u1ΓCu4)(u3ΓDu2). (3.58)

Again, the traces vanish if ΓCγµ 6= C ∝ ΓDγµ with C a commuting number, which implies

that ΓC = ΓD. That is,

(u1γµu2)(u3γµu4) =

∑C

tr (ΓCγµΓCγµ)

16(u1ΓCu4)(u3ΓCu2)

= 14

[4(u1u4)(u3u2)− 2(u1γ

µu4)(u3γµu2)

− 2(u1γ5γµu4)(u3γ

5γµu2)− 4(u1γ5u4)(u3γ

5u2)]. (3.59)

We note that the normalization of Dirac matrices has been properly taken into account by

raising or lowering of Lorentz indices.

3.7 The discrete symmetries P , C and T

(a) In this problem, we will work out the C, P and T transformations of the bilinear

ψσµνψ, with σµν = i2[γµ, γν ]. Firstly,

Pψ(t,x)σµνψ(t,x)P = i2ψ(t,−x)γ0[γµ, γν ]γ0ψ(t,−x).

With the relations γ0[γ0, γi]γ0 = −[γ0, γi] and γ0[γi, γj]γ0 = [γi, γj], we get:

Pψ(t,x)σµνψ(t,x)P =

− ψ(t,−x)σ0iψ(t,−x);

ψ(t,−x)σijψ(t,−x).(3.60)

Secondly,

T ψ(t,x)σµνψ(t,x)T = − i2ψ(−t,x)(−γ1γ3)[γµ, γν ]∗(γ1γ3)ψ(−t,x).


Note that gamma matrices keep invariant under transposition, except γ2, which changes the

sign. Thus we have:

T ψ(t,x)σµνψ(t,x)T =

ψ(−t,x)σ0iψ(−t,x);

− ψ(−t,x)σijψ(−t,x).(3.61)

Thirdly,

Cψ(t,x)σµνψ(t,x)C = − i2(−iγ0γ2ψ)Tσµν(−iψγ0γ2)T = ψγ0γ2(σµν)Tγ0γ2ψ.

Note that γ0 and γ2 are symmetric while γ1 and γ3 are antisymmetric, we have

Cψ(t,x)σµνψ(t,x)C = −ψ(t,x)σµνψ(t,x). (3.62)

(b) Now we work out the C, P and T transformation properties of a scalar field φ. Our

starting point is

PapP = a−p, TapT = a−p, CapC = bp.

Then, for a complex scalar field

φ(x) =

∫d3k

(2π)3

1√2k0

[ake

−ik·x + b†keik·x], (3.63)

we have

Pφ(t,x)P =

∫d3k

(2π)3

1√2k0

[a−ke

−i(k0t−k·x) + b†−kei(k0t−k·x)

]= φ(t,−x). (3.64a)

Tφ(t,x)T =

∫d3k

(2π)3

1√2k0

[a−ke

i(k0t−k·x) + b†−ke−i(k0t−k·x)

]= φ(−t,x). (3.64b)

Cφ(t,x)C =

∫d3k

(2π)3

1√2k0

[bke−i(k0t−k·x) + a†ke

i(k0t−k·x)]

= φ∗(t,x). (3.64c)

As a consequence, we can deduce the C, P , and T transformation properties of the current

Jµ = i(φ∗∂µφ− (∂µφ∗)φ

), as follows:

PJµ(t,x)P = (−1)s(µ)i[φ∗(t,−x)∂µφ(t,−x)−

(∂µφ∗(t,−x)

)φ(t,−x)

]= (−1)s(µ)Jµ(t,−x), (3.65a)

where s(µ) is the label for space-time indices that equals to 0 when µ = 0 and 1 when

µ = 1, 2, 3. In the similar way, we have

TJµ(t,x)T = (−1)s(µ)Jµ(−t,x); (3.65b)

CJµ(t,x)C = −Jµ(t,x). (3.65c)

One should be careful when playing with T — it is antihermitian rather than hermitian,

and anticommutes, rather than commutes, with√−1.

3.8. Bound states 21

(c) Any Lorentz-scalar hermitian local operator O(x) constructed from ψ(x) and φ(x)

can be decomposed into groups, each of which is a Lorentz-tensor hermitian operator and

contains either ψ(x) or φ(x) only. Thus to prove that O(x) is an operator of CPT = +1, it

is enough to show that all Lorentz-tensor hermitian operators constructed from either ψ(x)

or φ(x) have correct CPT value. For operators constructed from ψ(x), this has been done

as listed in Table on Page 71 of Peskin & Schroeder; and for operators constructed from

φ(x), we note that all such operators can be decomposed further into a product (including

Lorentz inner product) of operators of the form

(∂µ1 · · · ∂µmφ†)(∂µ1 · · · ∂µnφ) + c.c

together with the metric tensor ηµν . But it is easy to show that any operator of this form

has the correct CPT value, namely, has the same CPT value as a Lorentz tensor of rank

(m+n). Therefore we conclude that any Lorentz-scalar hermitian local operator constructed

from ψ and φ has CPT = +1.

3.8 Bound states

(a) A positronium bound state with orbital angular momentum L and total spin S can

be build by linear superposition of an electron state and a positron state, with the spatial

wave function ΨL(k) as the amplitude. Symbolically we have

|L, S〉 ∼∑k

ΨL(k)a†(k, s)b†(−k, s′)|0〉.

Then, apply the space-inversion operator P , we get

P |L, S〉 =∑k

ΨL(−k)ηaηba†(−k, s)b†(k, s′)|0〉 = (−1)Lηaηb

∑k

ΨL(k)a†(k, s)b†(k, s′)|0〉.

(3.66)

Note that ηb = −η∗a, we conclude that P |L, S〉 = (−)L+1|L, S〉. Similarly,

C|L, S〉 =∑k

ΨL(k)b†(k, s)a†(−k, s′)|0〉 = (−1)L+S∑k

ΨL(k)b†(−k, s′)a†(k, s)|0〉. (3.67)

That is, C|L, S〉 = (−1)L+S|L, S〉. Then its easy to find the P and C eigenvalues of various

states, listed as follows:

SL 1S 3S 1P 3P 1D 3D

P − − + + − −C + − − + + −

(b) We know that a photon has parity eigenvalue −1 and C-eigenvalue −1. Thus we see

that the decay into 2 photons are allowed for 1S state but forbidden for 3S state due to

C-violation. That is, 3S has to decay into at least 3 photons.


Chapter 4

Interacting Fields and Feynman

Diagrams

4.1 Scalar field with a classical source

In this problem we consider the theory with the following Hamiltonian:

H = H0 −∫

d3 j(t,x)φ(x), (4.1)

where H0 is the Hamiltonian for free Klein-Gordon field φ, and j is a classical source.

(a) We calculate the probability that the source creates no particles. The corresponding

amplitude is given by the inner product between the in-state and the out-state, both of

which are vacuum in our case. Therefore,

P (0) =∣∣out〈0|0〉in

∣∣2 = limt→(1−iε)∞

∣∣〈0|e−i2Ht|0〉∣∣2=∣∣∣〈0|T exp

− i∫

d4xHint

|0〉∣∣∣2 =

∣∣∣〈0|T expi

∫d4x j(x)φI(x)

|0〉∣∣∣2. (4.2)

(b) Now we expand this probability P (0) to j2. The amplitude reads,

〈0|T expi

∫d4x j(x)φI(x)

|0〉 =1− 1

2

∫d4x d4y j(x)〈0|TφI(x)φI(y)|0〉j(y) +O(j4)

=1− 1

2

∫d4x d4y j(x)j(y)

∫d3p

(2π)3

1

2Ep

+O(j4)

=1− 1

2

∫d3p

(2π)3

1

2Ep

|j(p)|2 +O(j4). (4.3)

Thus the probability is given by,

P (0) = |1− 1

2λ+O(j4)|2 = 1− λ+O(j4), (4.4)

where

λ ≡∫

d3p

(2π)3

1

2Ep

|j(p)|2. (4.5)

23

24 Chapter 4. Interacting Fields and Feynman Diagrams

(c) We can calculate the probability P (0) exactly, by working out the j2n term of the

expansion as,

i2n

(2n)!

∫d4x1 · · · d4x2n j(x1) · · · j(x2n)〈0|Tφ(x1) · · ·φ(x2n)|0〉

=i2n(2n− 1)(2n− 3) · · · 3 · 1

(2n)!

∫d4x1 · · · d4x2n j(x1) · · · j(x2n)∫

d3p1 · · · d3pn(2π)3n

1

2nEp1 · · ·Epn

eip1·(x1−x2) · · · eipn·(x2n−1−x2n)

=(−1)n

2nn!

(∫ d3p

(2π)3

|j(p)|2

2Ep

)n=

(−λ/2)2

n!. (4.6)

Then,

P (0) =( ∞∑n=0

(−λ/2)n

n!

)2

= e−λ. (4.7)

(d) Now we calculate the probability that the source creates one particle with momentum

k, which is given by,

P (k) =∣∣∣〈k|T exp

i

∫d4x j(x)φI(x)

|0〉∣∣∣2 (4.8)

Expanding the amplitude to the first order in j, we get:

P (k) =∣∣∣〈k|0〉+ i

∫d4x j(x)

∫d3p

(2π)3

eip·x√2Ep

〈k|a†p|0〉+O(j2)∣∣∣2

=∣∣∣i ∫ d3p

(2π)3

j(p)√2Ep

√2Ep(2π)3δ(p− k)

∣∣∣2 = |j(k)|2 +O(j3). (4.9)

If we go on to work out all the terms, we get,

P (k) =∣∣∣∑

n

i(2n+ 1)(2n+ 1)(2n− 1) · · · 3 · 1(2n+ 1)!

jn+1(k)∣∣∣2 = |j(k)|2e−|j(k)|. (4.10)

(e) To calculate the probability that the source creates n particles, we write down the

relevant amplitude,∫d3k1 · · · d3kn

(2π)3n√

2nEk1 · · ·Ekn

〈k1 · · ·kn|T expi

∫d4x j(x)φI(x)

|0〉. (4.11)

Expanding this amplitude in terms of j, we find that the first nonvanishing term is the one

of n’th order in j. Repeat the similar calculations above, we can find that the amplitude is:

in

n!

∫d3k1 · · · d3kn

(2π)3n√

2nEk1 · · ·Ekn

∫d4x1 · · · d4xn j1 · · · jn〈k1 · · ·kn|φ1 · · ·φn|0〉+O(jn+2)

=in

n!

∫d3k1 · · · d3kn j

n(k)

(2π)3n√

2nEk1 · · ·Ekn

∞∑n=0

(−1)n

2nn!

(∫ d3p

(2π)3

|j(p)|2

2Ep

)n. (4.12)

4.2. Decay of a scalar particle 25

Then we see the probability is given by,

P (n) =λn

n!e−λ, (4.13)

which is a Poisson distribution.

(f) It’s easy to check that the Poisson distribution P (n) satisfies the following identities:∑n

P (n) = 1. (4.14)

〈N〉 =∑n

nP (n) = λ. (4.15)

The first one is almost trivial, and the second one can be obtained by acting λ ddλ

to both

sides of the first identity. If we apply λ ddλ

again to the second identity, we get:

〈(N − 〈N〉)2〉 = 〈N2〉 − 〈N〉2 = λ. (4.16)

4.2 Decay of a scalar particle

This problem is based on the following Lagrangian,

L =1

2(∂µΦ)2 − 1

2M2Φ2 +

1

2(∂µφ)2 − 1

2m2φ2 − µΦφφ. (4.17)

When M > 2m, a Φ particle can decay into two φ particles. We want to calculate the

lifetime of the Φ particle to lowest order in µ.

The two-body decay rate is given in (4.86) of P&S,∫dΓ =

1

2M

∫d3p1d3p2

(2π)6

1

4Ep1Ep2

∣∣M(Φ(0)→ φ(p1)φ(p2))∣∣2(2π)4δ(4)(pΦ−p1−p2). (4.18)

To lowest order in µ, the amplitude M is given by,

iM = −2iµ. (4.19)

The delta function in our case reads,

δ(4)(pΦ − p1 − p2) = δ(M − Ep1 − Ep2)δ(3)(p1 + p2), (4.20)

thus,

Γ =1

2· 2µ2

M

∫d3p1d3p2

(2π)6

1

4Ep1Ep2

(2π)4δ(M − Ep1 − Ep2)δ(3)(p1 + p2), (4.21)

where an additional factor of 1/2 takes account of two identical φ’s in final state. Further-

more, there are two mass-shell constraints,

m2 + p2i = E2

pi. (i = 1, 2) (4.22)


Hence,

Γ =µ2

M

∫d3p1

(2π)3

1

4E2p1

(2π)δ(M − 2Ep1) =µ2

8πM

(1− 4m2

M2

)1/2

. (4.23)

Then the lifetime τ of Φ is,

τ = Γ−1 =8πM

µ2

(1− 4m2

M2

)−1/2

. (4.24)

4.3 Linear sigma model

In this problem, we study the linear sigma model described by the following Lagrangian,

L = 12∂µΦi∂µΦi − 1

2m2ΦiΦi − 1

4λ(ΦiΦi)2. (4.25)

Where Φ is an N -component scalar.

(a) We firstly compute the differential cross sections to the leading order in λ for the

following three processes,

Φ1Φ2 → Φ1Φ2, Φ1Φ1 → Φ2Φ2, Φ1Φ1 → Φ1Φ1. (4.26)

Since the masses of all incoming and outgoing particles are identical, the cross section is

simply given by ( dσ

dΩ

)CM

=|M|2

64π2s, (4.27)

where s is the square of center-of-mass energy, and M is the scattering amplitude. From

the Feynman rules it’s easy to get,

M(Φ1Φ2 → Φ1Φ2) =M(Φ1Φ1 → Φ2Φ2) = −2iλ,

M(Φ1Φ1 → Φ1Φ1) = −6iλ. (4.28)

It follows immediately that

σ(Φ1Φ2 → Φ1Φ2) = σM(Φ1Φ1 → Φ2Φ2) =λ2

16π2s,

σ(Φ1Φ1 → Φ1Φ1) =9λ2

16π2s. (4.29)

(b) Now we study the symmetry broken case, that is, m2 = −µ2 < 0. Then, the scalar

multiplet Φ can be parameterized as

Φ = (π1, · · · , πN−1, σ + v)T , (4.30)

where v is the VEV of |Φ|, and equals to√µ2/λ at tree level.

Substitute this into the Lagrangian, we get

L = 12

(∂µπk)2 + 1

2(∂µσ)2 − 1

2(2µ2)σ2 −

√λµσ3 −

√λµσπkπk

4.3. Linear sigma model 27

− λ4σ4 − λ

2σ2(πkπk)− λ

4(πkπk)2. (4.31)

Then it’s easy to read the Feynman rules from this expression:

k=

i

k2 − 2µ2; (4.32a)

k=

iδij

k2; (4.32b)

= 6iλv; (4.32c)

i j

= − 2iλvδij; (4.32d)

= − 6iλ; (4.32e)

i j

= − 2iλδij; (4.32f)

i j

` k

= − 2iλ(δijδk` + δikδj` + δi`δjk). (4.32g)

(c) With the Feynman rules derived in (b), we can compute the amplitude

M[πi(p1)πj(p2)→ πk(p3)π`(p4)

],

as:

M = (−2iλv)2[ i

s− 2µ2δijδk` +

i

t− 2µ2δikδj` +

i

u− 2µ2δi`δjk

]− 2iλ(δijδk` + δikδj` + δi`δjk), (4.33)

where s, t, u are Mandelstam variables (See Section 5.4 of P&S). Then, at the threshold

pi = 0, we have s = t = u = 0, and M vanishes.

On the other hand, if N = 2, then there is only one component in π, thus the amplitude

reduces to

M =− 2iλ[ 2µ2

s− 2µ2+

2µ2

t− 2µ2+

2µ2

u− 2µ2+ 3]


= 2iλ[ s+ t+ u

2µ2+O(p4)

]. (4.34)

In the second line we perform the Taylor expansion on s, t and u, which are of order O(p2).

Note that s+ t+ u = 4m2π = 0, thus we see that O(p2) terms are also canceled out.

(d) We minimize the potential with a small symmetry breaking term:

V = −µ2ΦiΦi + λ4

(ΦiΦi)2 − aΦN , (4.35)

which yields the following equation that determines the VEV:(− µ2 + λΦiΦi

)Φi = aδiN . (4.36)

Thus, up to linear order in a, the VEV 〈Φi〉 = (0, · · · , 0, v) is

v =

√µ2

λ+

a

2µ2. (4.37)

Now we repeat the derivation in (b) with this new VEV, and write the Lagrangian in terms

of new field variable πi and σ, as

L = 12

(∂µπk)2 + 1

2(∂µσ)2 − 1

2

√λaµπkπk − 1

2(2µ2)σ2

− λvσ3 − λvσπkπk − 14λσ4 − λ

2σ2(πkπk)− λ

4(πkπk)2. (4.38)

The πiπj → πkπ` amplitude is still given by

M = (−2iλv)2[ i

s− 2µ2δijδk` +

i

t− 2µ2δikδj` +

i

u− 2µ2δi`δjk

]− 2iλ(δijδk` + δikδj` + δi`δjk). (4.39)

However this amplitude does not vanish at the threshold. Since the vertices λν 6=√λµ

exactly even at tree level, and also s, t and u are not exactly zero in this case due to nonzero

mass of πi. Both deviations are proportional to a, thus we conclude that the amplitude Mis also proportional to a.

4.4 Rutherford scattering

The Rutherford scattering is the scattering of an election by the coulomb field of a

nucleus. In this problem, we calculate the cross section by treating the electromagnetic field

as fixed classical background given by potential Aµ(x). Then the interaction Hamiltonian

is,

HI =

∫d3x eψγµψAµ. (4.40)

4.4. Rutherford scattering 29

(a) We first calculate the T -matrix to lowest order,

out〈p′|p〉in =〈p′|T exp(−i∫

d4xHI)|p〉 = 〈p′|p〉 − ie∫

d4xAµ(x)〈p′|ψγµψ|p〉+O(e2)

=〈p′|p〉 − ie∫

d4xAµ(x)u(p′)γµu(p)ei(p′−p)·x +O(e2)

=(2π)4δ(4)(p− p′)− ieu(p′)γµu(p)Aµ(p′ − p) +O(e2). (4.41)

On the other hand,

out〈p′|p〉in = 〈p′|S|p〉 = 〈p′|p〉+ 〈p′|iT |p〉. (4.42)

Thus to the first order of e, we get,

〈p′|iT |p〉 = −ieu(p′)γµu(p)Aµ(p′ − p). (4.43)

(b) Now we calculate the cross section dσ in terms of the matrix elements iM.

The incident wave packet |ψ〉 is defined to be:

|ψ〉 =

∫d3k

(2π)3

e−ib·k√2Ek

ψ(k)|k〉, (4.44)

where b is the impact parameter.

The probability that a scattered electron will be found within an infinitesimal element

d3p centered at p is,

P =d3p

(2π)3

1

2Ep

∣∣∣out〈p|ψ〉in∣∣∣2

=d3p

(2π)3

1

2Ep

∫d3kd3k′

(2π)6√

2Ek2Ek′ψ(k)ψ∗(k′)

(out〈p|k〉in

)(out〈p|k′〉in

)∗e−ib·(k−k

′)

=d3p

(2π)3

1

2Ep

∫d3kd3k′

(2π)6√

2Ek2Ek′ψ(k)ψ∗(k′)

(〈p|iT |k〉

)(〈p|iT |k′〉

)∗e−ib·(k−k

′). (4.45)

In the last equality we throw away the trivial scattering part from the S-matrix. Note that,

〈p′|iT |p〉 = iM(2π)δ(Ep′ − Ep), (4.46)

so we have,

P =d3p

(2π)3

1

2Ep

∫d3kd3k′

(2π)6√

2Ek2Ek′ψ(k)ψ∗(k′)|iM|2(2π)2δ(Ep − Ek)δ(Ep − Ek′)e

−ib·(k−k′).

(4.47)

The cross section dσ is given by:

dσ =

∫d2b P (b), (4.48)

thus the integration over b gives a delta function:∫d2b e−ib·(k−k

′) = (2π)2δ(2)(k⊥ − k′⊥). (4.49)


The other two delta functions in the integrand can be modified as follows,

δ(Ek − Ek′) =Ek

k‖δ(k‖ − k′‖) =

1

vδ(k‖ − k′‖), (4.50)

where we have used |v| = v = v‖. Taking all these delta functions into account, we get,

dσ =d3p

(2π)3

1

2Ep

∫d3k

(2π)32Ek

1

vψ(k)ψ∗(k)|iM|2(2π)δ(Ep − Ek). (4.51)

Since the momentum of the wave packet should be localized around its central value, we

can pull out the quantities involving energy Ek outside the integral,

dσ =d3p

(2π)3

1

2Ep

1

2Ek

1

v(2π)|M|2δ(Ep − Ek)

∫d3k

(2π)3ψ(k)ψ∗(k). (4.52)

Recall the normalization of the wave packet,∫d3k

(2π)3ψ(k)∗ψ(k) = 1, (4.53)

then,

dσ =d3p

(2π)3

1

2Ep

1

2Ek

1

v|M(k → p)|2(2π)δ(Ep − Ek). (4.54)

We can further integrate over |p| to get the differential cross section dσ/dΩ,

dσ

dΩ=

∫dp p2

(2π)3

1

2Ep

1

2Ek

1

v|M(k → p)|2(2π)δ(Ep − Ek)

=

∫dp p2

(2π)3

1

2Ep

1

2Ek

1

v|M(k → p)|2(2π)

Ek

kδ(p− k)

=1

(4π)2|M(k, θ)|2. (4.55)

In the last line we work out the integral by virtue of delta function, which constrains the

outgoing momentum |p| = |k| but leave the angle θ between p and k arbitrary. Thus the

amplitude M(k, θ) is a function of momentum |k| and angle θ.

(c) We work directly for the relativistic case. Firstly the Coulomb potential A0 = Ze/4πr

in momentum space is

A0(q) =Ze

|q|2. (4.56)

This can be easily worked out by Fourier transformation, with a “regulator” e−mr inserted:

A0(q,m) ≡∫

d3x e−ip·xe−mrZe

4πr=

Ze

|q|2 +m2. (4.57)

This is simply Yukawa potential, and Coulomb potential is a limiting case when m→ 0.

The amplitude is given by

iM(k, θ) = ieu(p)γµAµ(q)u(p) with q = p− k. (4.58)

4.4. Rutherford scattering 31

Then we have the squared amplitude with initial spin averaged and final spin summed (See

§5.1 of P&S for details), as,

12

∑spin

|iM(k, θ)|2 = 12e2Aµ(q)Aν(q)

∑spin

u(p)γµu(k)u(k)γνu(p)

= 12e2Aµ(q)Aν(q) tr

[γµ(/p+m)γν(/k +m)

]=2e2

[2(p · A)(k · A) +

(m2 − (k · p)

)A2]. (4.59)

Note that

A0(q) =Ze

|p− k|2=

Ze

4|k|2 sin2(θ/2), (4.60)

thus

12

∑spin

|iM(k, θ)|2 =Z2e4

(1− v2 sin2 θ

2

)4|k|4v2 sin4(θ/2)

, (4.61)

anddσ

dΩ=

Z2α2(1− v2 sin2 θ

2

)4|k|2v2 sin4(θ/2)

(4.62)

. In non-relativistic case, this formula reduces to

dσ

dΩ=

Z2α2

4m2v4 sin4(θ/2)(4.63)


Chapter 5

Elementary Processes of Quantum

Electrodynamics

5.1 Coulomb scattering

In this problem we continue our study of the Coulomb scattering in Problem 4.4. Here

we consider the relativistic case. Let’s first recall some main points considered before. The

Coulomb potential A0 = Ze/4πr in momentum space is

A0(q) =Ze

|q|2. (5.1)

Then the scattering amplitude is given by

iM(k, θ) = ieu(p)γµAµ(q)u(p) with q = p− k. (5.2)

Then we can derive the squared amplitude with initial spin averaged and final spin summed,

as:

1

2

∑spin

|iM(k, θ)|2 =1

2e2Aµ(q)Aν(q)

∑spin

u(p)γµu(k)u(k)γνu(p)

=1

2e2Aµ(q)Aν(q) tr

[γµ(/p+m)γν(/k +m)

]= 2e2

[2(p · A)(k · A) +

(m2 − (k · p)

)A2]. (5.3)

Note that

A0(q) =Ze

|p− k|2=

Ze

4|k|2 sin2(θ/2), (5.4)

thus1

2

∑spin

|iM(k, θ)|2 =Z2e4

(1− v2 sin2 θ

2

)4|k|4v2 sin4(θ/2)

, (5.5)

Now, from the result of Problem 4.4(b), we know that

dσ

dΩ=

1

(4π)2

(12

∑spin

|M(k, θ)|2)

=Z2α2

(1− v2 sin2 θ

2

)4|k|2v2 sin4(θ/2)

. (5.6)

33

34 Chapter 5. Elementary Processes of Quantum Electrodynamics

This is the formula for relativistic electron scatted by Coulomb potential, and is called Mott

formula.

Now we give an alternative derivation of the Mott formula, by considering the cross

section of e−µ−Z → e−µ−Z . When the mass of µ goes to infinity and the charge of µ is

taken to be Ze, this cross section will reduces to Mott formula.

k1 k2

p1 p2

e− µ−Z

e− µ−Z

Figure 5.1: The scattering of an electron by a charged heavy particle µ−Z . All initial

momenta go inward and all final momenta go outward.

The corresponding Feynman diagram is shown in Figure 5.1, which reads,

iM = Z(−ie)2u(p1)γµu(k1)−i

tU(p2)γµU(k2), (5.7)

where u is the spinor for electron and U is the spinor for muon, t = (k1−p1)2 is one of three

Mandelstam variables. Then the squared amplitude with initial spin averaged and final spin

summed is

14

∑spin

|iM|2 =Z2e4

t2tr[γµ(/k1 +m)γν(/p1

+m)]

tr[γµ(/k2 +M)γν(/p2

+M)]

=Z2e4

t2

[16m2M2 − 8M2(k1 · p1) + 8(k1 · p2)(k2 · p1)

− 8m2(k2 · p2) + 8(k1 · k2)(p1 · p2)]. (5.8)

Note that the cross section is given by( dσ

dΩ

)CM

=1

2Ee2Eµ|vk1 − vk2||p1|

(2π)24ECM

(14

∑|M|2

). (5.9)

When the mass of µ goes to infinity, we have Eµ ' ECM ' M , vk2 ' 0, and |p1| ' |k1|.Then the expression above can be simplified to( dσ

dΩ

)CM

=1

16(2π)2M2

(14

∑|M|2

). (5.10)

When M → ∞, only terms proportional to M2 are relevant in |M|2. To evaluate this

squared amplitude further, we assign each momentum a specific value in CM frame,

k1 = (E, 0, 0, k), p1 ' (E, sin θ, 0, k cos θ),

k2 ' (M, 0, 0,−k), p2 ' (M,−k sin θ, 0,−k cos θ), (5.11)

5.2. Bhabha scattering 35

then t = (k1 − p1)2 = 4k2 sin2 θ2

, and,

14

∑|iM|2 =

Z2e4(1− v2 sin2 θ2

)

k2v2 sin2 θ2

M2 +O(M). (5.12)

Substituting this into the cross section, and sending M → ∞, we reach the Mott formula

again,dσ

dΩ=

Z2α2(1− v2 sin2 θ

2

)4|k|2v2 sin4(θ/2)

. (5.13)

5.2 Bhabha scattering

The Bhabha scattering is the process e+e− → e+e−. At the tree level, it consists of two

diagrams, as shown in Figure 5.2.

k1

p2

k2

p1

e− e+

e− e+

−

k1

p2p1

k2

e− e+

e− e+

Figure 5.2: Bhabha scattering at tree level. All initial momenta go inward and all final

momenta go outward.

The minus sign before the t-channel diagram comes from the exchange of two fermion field

operators when contracting with in and out states. In fact, the s- and t-channel diagrams

correspond to the following two ways of contraction, respectively,

〈p1p2|ψ /Aψψ /Aψ|k1k2〉, 〈p1p2|ψ /Aψψ /Aψ|k1k2〉. (5.14)

In the high energy limit, we can omit the mass of electrons, then the amplitude for the

whole scattering process is,

iM = (−ie)2

[v(k2)γµu(k1)

−i

su(p1)γµv(p2)− u(p1)γµu(k1)

−i

tv(k2)γµv(p2)

], (5.15)

where we have used the Mandelstam variables s, t and u. They are defined as,

s = (k1 + k2)2, t = (p1 − k1)2, u = (p2 − k1)2. (5.16)

In the massless case, k21 = k2

2 = p21 = p2

2 = 0, thus we have,

s = 2k1 · k2 = 2p1 · p2, t = −2p1 · k1 = −2p2 · k2, u = −2p2 · k1 = −2p1 · k2. (5.17)

We want to get the unpolarized cross section, thus we must average the ingoing spins and

sum over outgoing spins. That is,

1

4

∑spin

|M|2 =e4

4s2

∑∣∣∣v(k2)γµu(k1)u(p1)γµv(p2)∣∣∣2


+e4

4t2

∑∣∣∣u(p1)γµu(k1)v(k2)γµv(p2)∣∣∣2

− e4

4st

∑[v(p2)γµu(p1)u(k1)γµv(k2)u(p1)γνu(k1)v(k2)γνv(p2) + c.c.

]=

e4

4s2tr (/k1γ

µ/k2γν) tr (/p2

γµ/p1γν) +

e4

4t2tr (/k1γ

µ/p1γν) tr (/p2

γµ/k2γν)

− e4

4st

[tr (/k1γ

ν/k2γµ/p2γν/p1

γµ) + c.c.]

=2e4(u2 + t2)

s2+

2e4(u2 + s2)

t2+

4e4u2

st

= 2e4

[t2

s2+s2

t2+ u2

( 1

s+

1

t

)2]. (5.18)

In the center-of-mass frame, we have k01 = k0

2 ≡ k0, and k1 = −k2, thus the total energy

E2CM = (k0

1 + k02)2 = 4k2 = s. According to the formula for the cross section in the four

identical particles’ case (Eq.4.85):( dσ

dΩ

)CM

=1

64π2ECM

(14

∑|M|2

), (5.19)

thus ( dσ

dΩ

)CM

=α2

2s

[t2

s2+s2

t2+ u2

( 1

s+

1

t

)2], (5.20)

where α = e2/4π is the fine structure constant. We integrate this over the angle ϕ to get:( dσ

d cos θ

)CM

=πα2

s

[t2

s2+s2

t2+ u2

( 1

s+

1

t

)2]. (5.21)

5.3 The spinor products (2)

In this problem we continue our study of spinor product method in last chapter. The

formulae needed in the following are:

uL(p) =1√

2p · k0/puR0, uR(p) =

1√2p · k0

/puL0. (5.22)

s(p1, p2) = uR(p1)uL(p2), t(p1, p2) = uL(p1)uR(p2). (5.23)

For detailed explanation for these relations, see Problem 3.3.

(a) Firstly, we prove the following relation,

|s(p1, p2)|2 = 2p1 · p2. (5.24)

We make use of the another two relations,

uL0uL0 =1− γ5

2/k0, uR0uR0

1 + γ5

2/k0, (5.25)

5.3. The spinor products (2) 37

which are direct consequences of the familiar spin-sum formula∑u0u0 = /k0. We now

generalize this to,

uL(p)uL(p) =1− γ5

2/p, uR(p)uR(p) =

1 + γ5

2/p. (5.26)

We prove the first one:

uL(p)uL(p) =1

2p · k0/puR0uR0/p =

1

2p · k0/p

1 + γ5

2/k0/p

=1

2p · k0

1− γ5

2/p/k0/p =

1

2p · k0

1− γ5

2(2p · k − /k0/p)/p

=1− γ5

2/p−

1

2p · k0

1− γ5

2/k0p

2 =1− γ5

2/p. (5.27)

The last equality holds because p is lightlike. Then we get,

|s(p1, p2)|2 =|uR(p1)uL(p2)|2 = tr(uL(p2)uL(p2)uR(p1)uR(p1)

)=

1

4tr((1− γ5)/p2

(1− γ5)/p1

)= 2p1 · p2. (5.28)

(b) Now we prove the relation,

tr (γµ1γµ2 · · · γµn) = tr (γµn · · · γµ2γµ1), (5.29)

where µi = 0, 1, 2, 3, 5.

To make things easier, let us perform the proof in Weyl representation, without loss of

generality. Then it’s easy to check that

(γµ)T =

γµ, µ = 0, 2, 5;

− γµ, µ = 1, 3.(5.30)

Then, we define M = γ1γ3, and it can be easily shown that M−1γµM = (γµ)T , and M−1M =

1. Then we have,

tr (γµ1γµ2 · · · γµn) = tr (M−1γµ1MM−1γµ2M · · ·M−1γµnM)

= tr[(γµ1)T

(γµ2)T · · · (γµn)T

]= tr

[(γµn · · · γµ2γµ1)T

]= tr (γµn · · · γµ2γµ1). (5.31)

With this formula in hand, we can derive the equality,

uL(p1)γµuL(p2) = uR(p2)γµuR(p1), (5.32)

as follows,

LHS = CuR0/p1γµ/p2

uR0 = C tr (/p1γµ/p2

)

= C tr (/p2γµ/p1

) = CuL0/p2γµ/p1

uL0 = RHS,

in which C ≡(2√

(p1 · k0)(p2 · k0))−1

.


(c) The way of proving the Fierz identity

uL(p1)γµuL(p2)[γµ]ab = 2[uL(p2)uL(p1) + uR(p1)uR(p2)

]ab

(5.33)

has been indicated in P&S. The right hand side of this identity, as a Dirac matrix, which we

denoted by M , can be written as a linear combination of 16 Γ matrices listed in Problem

3.6. In addition, it is easy to check directly that γµM = −Mγ5. Thus M must have the

form

M =( 1− γ5

2

)γµV

µ +( 1 + γ5

2

)γµW

µ.

Each of the coefficients V µ and W µ can be determined by projecting out the other one with

the aid of trace technology, that is,

V µ =1

2tr[γµ( 1− γ5

2

)M]

= uL(p1)γµuL(p2), (5.34)

W µ =1

2tr[γµ( 1 + γ5

2

)M]

= uR(p2)γµuR(p1) = uL(p1)γµuL(p2). (5.35)

The last equality follows from (5.32). Substituting V µ and W µ back, we finally get the left

hand side of the Fierz identity, which completes the proof.

(d) The amplitude for the process at leading order in α is given by,

iM = (−ie2)uR(k2)γµuR(k1)−i

svR(p1)γµvR(p2). (5.36)

To make use of the Fierz identity, we multiply (5.33), with the momenta variables changed

to p1 → k1 and p2 → k2, by[vR(p1)

]a

and[vR(p2)

]b, and also take account of (5.32), which

leads to,

uR(k2)γµuR(k1)vR(p1)γµvR(p2)

= 2[vR(p1)uL(k2)uL(k1)vR(p2) + vR(p1)uR(k1)uR(k2)vR(p2)

]= 2s(p1, k2)t(k1, p2). (5.37)

Then,

|iM|2 =4e4

s2|s(p1, k2)|2|t(k1, p2)|2 =

16e4

s2(p1 · k2)(k1 · p2) = e4(1 + cos θ)2, (5.38)

anddσ

dΩ(e+Le−R → µ+

Lµ−R) =

|iM|2

64π2Ecm

=α2

4Ecm

(1 + cos θ)2. (5.39)

It is straightforward to work out the differential cross section for other polarized processes

in similar ways. For instance,

dσ

dΩ(e+Le−R → µ+

Rµ−L) =

e4|t(p1, k1)|2|s(k2, p2)|2

64π2Ecm

=α2

4Ecm

(1− cos θ)2. (5.40)

5.4. Positronium lifetimes 39

(e) Now we recalculate the Bhabha scattering studied in Problem 5.2, by evaluating all

the polarized amplitudes. For instance,

iM(e+Le−R → e+

Le−R)

= (−ie)2

[uR(k2)γµuR(k1)

−i

svR(p1)γµvR(p2)

− uR(p1)γµuR(k1)−i

tvR(k2)γµvR(p2)

]= 2ie2

[s(p1, k2)t(k1, p2)

s− s(k2, p1)t(k1, p2)

t

]. (5.41)

Similarly,

iM(e+Le−R → e+

Re−L) = 2ie2 t(p1, k1)s(k2, p2)

s, (5.42)

iM(e+Re−L → e+

Le−R) = 2ie2 s(p1, k1)t(k2, p2)

s, (5.43)

iM(e+Re−L → e+

Re−L) = 2ie2

[t(p1, k2)s(k1, p2)

s− t(k2, p1)s(k1, p2)

t

], (5.44)

iM(e+Re−R → e+

Re−R) = 2ie2 t(k2, k1)s(p1, p2)

t, (5.45)

iM(e+Le−L → e+

Le−L) = 2ie2 s(k2, k1)t(p1, p2)

t. (5.46)

Squaring the amplitudes and including the kinematic factors, we find the polarized differ-

ential cross sections as,

dσ

dΩ(e+Le−R → e+

Le−R) =

dσ

dΩ(e+Re−L → e+

Re−L) =

α2u2

2s

( 1

s+

1

t

)2

, (5.47)

dσ

dΩ(e+Le−R → e+

Re−L) =

dσ

dΩ(e+Re−L → e+

Le−R) =

α2

2s

t2

s2, (5.48)

dσ

dΩ(e+Re−R → e+

Re−R) =

dσ

dΩ(e+Le−L → e+

Le−L) =

α2

2s

s2

t2. (5.49)

Therefore we recover the result obtained in Problem 5.2,

dσ

dΩ(e+e− → e+e−) =

α2

2s

[t2

s2+s2

t2+ u2

( 1

s+

1

t

)2]. (5.50)

5.4 Positronium lifetimes

In this problem we study the decay of positronium (Ps) in its S and P states. To

begin with, we recall the formalism developed for bound states with nonrelativistic quantum

mechanics in P&S. The positronium state |Ps〉, as a bound state of an electron-positron pair,

can be represented in terms of electron and positron’s state vectors, as,

|Ps〉 =√

2MP

∫d3k

(2π)3ψ(k)Cab

1√2m|e−a (k)〉 1√

2m|e+b (−k)〉, (5.51)


where m is the electron’s mass, MP is the mass of the positronium, which can be taken to be

2m as a good approximation, a and b are spin labels, the coefficient Cab depends on the spin

configuration of |Ps〉, and ψ(k) is the momentum space wave function for the positronium

in nonrelativistic quantum mechanics. In real space, we have,

ψ100(r) =

√(αmr)

3

πexp(−αmrr), (5.52)

ψ21i(r) =

√(αmr/2)5

πxi exp(−αmrr/2). (5.53)

where mr = m/2 is the reduced mass. Then the amplitude of the decay process Ps → 2γ

can be represented in terms of the amplitude for the process e+e− → 2γ as,

M(Ps→ 2γ) =1√m

∫d3k

(2π)3ψ(k)CabM

(e−a (k)e+

b (−k)→ 2γ). (5.54)

We put a hat on the amplitude of e+e− → 2γ. In the following we always use a hat to

denote the amplitude of this process.

(a) In this part we study the decay of the S-state positronium. As stated above, we have

to know the amplitude of the process e+e− → 2γ, which is illustrated in Figure 5.3 with the

B replaced with γ, and is given by,

iM = (−ie)2ε∗µ(p1)ε∗ν(p2)

× v(k2)

[γν

i(/k1 − /p1+m)

(k1 − p1)2 −m2γµ + γµ

i(/k1 − /p2+m)

(k1 − p2)2 −m2γν]u(k1), (5.55)

where the spinors can be written in terms of two-component spinors ξ and ξ′ in the chiral

representation as,

u(k1) =

(√k1 · σξ√k1 · σξ

), v(k2) =

( √k2 · σξ′

−√k2 · σξ′

). (5.56)

We also write γµ as,

γµ =

(0 σµ

σµ 0

),

where σµ = (1, σi) and σµ = (1,−σi) with σi the three Pauli matrices. Then the amplitude

can be brought into the following form,

iM = −ie2ε∗µ(p1)ε∗ν(p2)ξ′†[

Γµνt(k1 − p1)2 −m2

+Γµνu

(k1 − p2)2 −m2

]ξ, (5.57)

with

Γµνt =(√

k2 · σσνσµ√k1 · σ −

√k2 · σσν σµ

√k1 · σ

)m

+(√

k2 · σσνσλσµ√k1 · σ −

√k2 · σσν σλσµ

√k1 · σ

)(k1 − p1)λ,


Γµνu =(√

k2 · σσµσν√k1 · σ −

√k2 · σσµσν

√k1 · σ

)m

+(√

k2 · σσµσλσν√k1 · σ −

√k2 · σσµσλσν

√k1 · σ

)(k1 − p2)λ.

In the rest of the part (a), we take the nonrelativistic limit, with the momenta chosen to be

kµ1 = kµ2 = (m, 0, 0, 0), pµ1 = (m, 0, 0,m), pµ2 = (m, 0, 0,−m). (5.58)

Accordingly, we can assign the polarization vectors for final photons to be,

εµ±(p1) = 1√2(0, 1,±i, 0), εµ±(p2) = 1√

2(0,−1,±i, 0). (5.59)

Now substituting the momenta (5.58) into (5.57), noticing that√ki · σ =

√ki · σ =

√m (i =

1, 2), and (k1 − p1)2 = (k1 − p2)2 = −m2, and also using the trick that one can freely make

the substitution σµ → −σµ since the temporal component of the polarization vectors εµalways vanishes, we get a much more simplified expression,

iM = ie2ε∗µ(p1)ε∗ν(p2)ξ′†(σνσ3σµ − σµσ3σν

)ξ. (5.60)

The positronium can lie in spin-0 (singlet) state or spin-1 (triplit) state. In the former

case, we specify the polarizations of final photons in all possible ways, and also make the

substitution ξξ′† → 1√2

(See (5.49) of P&S), which leads to,

iMs++ = −iMs

−− = i2√

2e2, iMs+− = iMs

−+ = 0, (5.61)

where the subscripts denote final photons’ polarizations and s means singlet. We show the

mid-step for calculating iMs++ as an example,

iMs++ =

ie2

2tr[(

(σ1 + iσ2)σ3(−σ1 + iσ2)− (−σ1 + iσ2)σ3(σ1 + iσ2))ξξ′†]

= i2√

2e2.

In the same way, we can calculate the case of triplet initial state. This time, we make the

substitution ξξ′† → n · σ/√

2, with n = (x ± iy)/√

2 or n = z, corresponding to three

independent polarizations. But it is straightforward to show that the amplitudes with these

initial polarizations all vanish, which is consistent with our earlier results by using symmetry

arguments in Problem 3.8.

Therefore it is enough to consider the singlet state only. The amplitude for the decay of

a positronium in its 1S0 state into 2γ then follows directly from (5.54), as

M±±(1S0 → 2γ) =ψ(x = 0)√

mMs±±, (5.62)

where ψ(x = 0) =√

(mα/2)3/π according to (5.52). Then the squared amplitude with final

photons’ polarizations summed is∑spin

∣∣M(1S0 → 2γ)∣∣2 =

|ψ(0)|2

2m

(|Ms

++|2 + |Ms−−|2

)= 16πα5m2. (5.63)


Finally we find the decay width of the process Ps(1S0)→ 2γ, to be

Γ(1S0 → 2γ) =1

2

1

4m

∫d3p1d3p2

(2π)62E12E2

∑∣∣M(1S0 → 2γ)∣∣2(2π)4δ(4)(pPs − p1 − p2)

=1

2

1

4m

∫d3p1

(2π)34m2

∑∣∣M(1S0 → 2γ)∣∣2(2π)δ(m− E1)

=1

2α5m, (5.64)

where an additional factor of 1/2 follows from the fact that the two photons in the final

state are identical particles.

(b) To study the decay of P state (l = 1) positronium, we should keep one power of

3-momenta of initial electron and positron. Thus we set the momenta of initial and final

particles, and also the polarization vectors of the latter, in e−e+ → 2γ, to be

kµ1 = (E, 0, 0, k), kµ2 = (E, 0, 0,−k),

pµ1 = (E,E sin θ, 0, E cos θ), pµ2 = (E,−E sin θ, 0,−E cos θ),

εµ±(p1) = 1√2(0, cos θ,±i,− sin θ), εµ±(p2) = 1√

2(0,− cos θ,±i, sin θ). (5.65)

Here we have the approximate expression up to linear order in k:√k1 · σ =

√k2 · σ =

√m− k

2√mσ3 +O(k2),√

k2 · σ =√k1 · σ =

√m+

k

2√mσ3 +O(k2),

1

(k1 − p1)2 −m2= − 1

2m2− k cos θ

2m3+O(k2),

1

(k1 − p2)2 −m2= − 1

2m2+k cos θ

2m3+O(k2).

Consequently,

Γµνt = 2m2σν(σ1sθ + σ3cθ)σµ −mk

(σ3σνσµ + σνσµσ3 + 2σνσ3σµ

)+O(k2),

Γµνu = −2m2σµ(σ1sθ + σ3cθ)σν −mk

(σ3σµσν + σµσνσ3 + 2σµσ3σν

)+O(k2),

where we use the shorthand notation sθ = sin θ and cθ = cos θ. We can use these expansion

to find the terms in the amplitude iM of linear order in k, to be

iM∣∣O(k)

=− ie2ε∗µ(p1)ε∗ν(p2)k

2mξ′†[− 2cθσ

µ(σ1sθ + σ3cθ)σν − 2cθσ

ν(σ1sθ + σ3cθ)σµ

+(σ3σµσν + σµσνσ3 + 2σµσ3σν

)+(σ3σνσµ + σνσµσ3 + 2σνσ3σµ

)]ξ, (5.66)

Feeding in the polarization vectors of photons, and also make the substitution ξξ′† → n ·σ/√

2 or 1/√

2 for triplet and singlet positronium, respectively, as done in last part, we get

iM↓↓±±|O(k) = 0, iM↓↓

±∓|O(k) = −i2sθ(∓1 + cθ)e2k/m,


iM↑↓+↓↑±± |O(k) = i2

√2e2k/m, iM↑↓+↓↑

±∓ |O(k) = i2√

2s2θe

2k/m,

iM↑↑±±|O(k) = 0, iM↑↑

±∓|O(k) = −i2sθ(±1 + cθ)e2k/m,

iM↑↓−↓↑±± |O(k) = 0, iM↑↓−↓↑

±∓ |O(k) = 0. (5.67)

The vanishing results in the last line indicate that S = 0 state of P -wave positronium cannot

decay to two photons.

(c) Now we prove that the state,

|B(k)〉 =√

2MP

∫d3p

(2π)3ψi(p)a†p+k/2Σib†−p+k/2|0〉 (5.68)

, is a properly normalized state for the P -wave positronium. In fact,

〈B(k)|B(k)〉 = 2MP

∫d3p′

(2π)3

d3p

(2π)3ψ∗j (p

′)ψi(p)

× 〈0|b−p′+k/2Σj†ap′+k/2a†p+k/2Σib†−p′+k/2|0〉

= 2MP

∫d3p′

(2π)3

d3p

(2π)3ψ∗j (p

′)ψi(p)

× 〈0|b−p′+k/2Σj†Σib†−p′+k/2|0〉(2π)3δ(3)(p′ − p)

= 2MP

∫d3p

(2π)3ψ∗j (p)ψi(p)〈0|b−p+k/2Σj†Σib†−p+k/2|0〉

= 2MP

∫d3p

(2π)3ψ∗j (p)ψi(p)〈0| tr (Σj†Σi)|0〉(2π)3δ(3)(0)

= 2MP · (2π)3δ(3)(0), (5.69)

which is precisely the needed normalization of a state. In this calculation we have used the

anticommutation relations of creation and annihilation operators, as well as the normaliza-

tion of the wave function and the Σ matrices.

(d) Now we evaluate the partial decay rate of the S = 1 P -wave positronium of definite

J into two photons. The states for the positronium is presented in (c), with the Σ matrices

chosen as

Σ =

1√6σi, J = 0,

1

2εijknjσk, J = 1,

1√3hijσj, J = 2,

(5.70)

and the wave function given by (5.53).

Firstly, consider the J = 0 state, in which case we have,

iM(3P0 → γαγβ) =1√m

∫d3k

(2π)3ψi(k)

( 1√6σi)ab

iM(e−a (k)e+

b (−k)→ γαγβ), (5.71)


where α, β = + or − are labels of photons’ polarizations and a, b =↑ or ↓ are spinor indices.

For amplitude iM, we only need the terms linear in k, as listed in (5.67). Let us rewrite

this as,

iM(e−a (k)e+

b (−k)→ γαγβ)

= F abαβ,ik

i.

In the same way, the wave function can also be put into the form of ψi(x) = xif(r), with

r = |x|. Then the integration above can be carried out to be,

iM(3P0 → γαγβ) =i√6m

σiabFabαβ,j

[i∂

∂xjψi(x)

]x=0

=i√6m

σiabFabαβ,if(0). (5.72)

On the other hand, we have chose the direction of k to be in the x3-axis, then F abαβ,1 =

F abαβ,2 = 0 as a consequence. Therefore,

iM(3P0 → γ±γ∓) =i√6m

f(0)(F ↑↑±∓,3 − F

↓↓±∓,3

)= ±

√πα7

24m sin θ. (5.73)

Square these amplitudes, sum over the photons’ polarizations, and finish the phase space

integration in the same way as what we did in (a), we finally get the partial decay rate of

the J = 0 P -wave positronium into two photons to be,

Γ(3P0) =1

576α7m. (5.74)

The positronium in 3P1 state, namely the case J = 1, cannot decay into two photons

by the conservation of the angular momentum, since the total angular momentum of two

physical photons cannot be 1. Therefore let us turn to the case of J = 2. In this case we

should average over the initial polarizations of the positronium, which can be represented

by the symmetric and traceless polarization tensors hijn , with n = 1, 2, · · · , 5 the labeled of

5 independent polarizations. Let us choose these tensors to be,

hij1 = 1√2(δi2δj3 + δi3δj2), hij2 = 1√

2(δi1δj3 + δi3δj1),

hij3 = 1√2(δi1δj2 + δi2δj1), hij4 = 1√

2(δi1δj1 − δi2δj2),

hij5 = 1√2(δi1δj1 − δi3δj3). (5.75)

Then the decay amplitude for a specific polarization of J = 2 Ps can be represented as,

iMn(3P2 → γαγβ) =1√m

∫d3k

(2π)3ψi(k)

( 1√3hijn σ

j)ab

iM(e−a (k)e+

b (−k)→ γαγβ)

=1√3m

hijn σjabF

abαβ,if(0). (5.76)

Now substituting all stuffs in, we find the nonvanishing components of the decay amplitude

to be,

iM2(3P2 → γ±γ±) =

√πα7

48im,

5.5. Physics of a massive vector boson 45

iM2(3P2 → γ±γ∓) =

√πα7

48im sin2 θ,

iM5(3P2 → γ±γ∓) = ∓2

√πα7

48im sin2 θ. (5.77)

Squaring these amplitudes, summing over photon’s polarizations and averaging the initial

polarization of the positronium (by dividing the squared and summed amplitude by 5), we

get,

1

5

∑spin

∣∣Mn(3P2 → 2γ)∣∣2 =

πα7m2

120(1 + sin2 θ + 4 sin4 θ). (5.78)

Finally, we finish the phase space integration and get the partial decay rate of 3P2 positro-

nium into 2 photons to be

Γ(3P2) =19

19200α7m. (5.79)

5.5 Physics of a massive vector boson

In this problem, the mass of electron is always set to zero.

(a) We firstly compute the cross section σ(e+e− → B) and the decay rate Γ(B → e+e−).

For the cross section, the squared amplitude can be easily found to be

1

4

∑spin

|iM|2 =1

4

∑spin

∣∣∣igε∗(i)µ v(p′)γµu(p)∣∣∣2 = 2g2(p · p′). (5.80)

Note that we have set the mass of electrons to be zero. Then the cross section can be

deduced from (4.79). Let’s take the initial momenta to be,

p = 12

(E, 0, 0, E), p′ = 12

(E, 0, 0,−E), (5.81)

with E being the center-of-mass energy. Then it’s easy to get,

σ =g2

4E(2π)δ(MB − E) =

g2

4E(2π)2MBδ(M

2B − s) = πg2δ(M2

B − s), (5.82)

where s = E2.

To deduce the decay rate, we should average polarizations of massive vector B instead

of two electrons. Thus the squared amplitude in this case reads,

1

3

∑spin

|iM|2 =8

3g2(p · p′). (5.83)

The decay rate can be found from (4.86),

Γ =1

2MB

∫d3p

(2π)3

d3p′

2π3

1

2Ep

1

2Ep′

(13

∑|M|2

)(2π)4δ(4)(pB − p− p′)


=1

2MB

∫d3p

(2π)3

1

4E2p

(163g2E2

p

)(2π)δ(MB − 2Ep)

=4π

(2π)22MB

∫dp 4

3g2E2

p12δ( 1

2MB − Ep) =

g2MB

12π. (5.84)

We see the cross section and the decay rate satisfy the following relation, as expected,

σ(e+e− → B) =12π2

MB

Γ(B → e+e−)δ(s−M2). (5.85)

(b) Now we calculate the cross section σ(e−e+ → γ + B) in COM frame. The related

diagrams are shown in Figure 5.3.

k1

p2p1

k2

e− e+

γ B

+

k1

p1

k2

p2

e− e+

γ B

Figure 5.3: The tree diagrams of the process e−e+ → γ+B. All initial momenta go inward

and all final momenta go outward.

The amplitude reads,

iM = (−ie)(−ig)ε∗µ(p1)e∗ν(p2)v(k2)[γν

i

/k1 − /p1

γµ + γµi

/k1 − /p2

γν]u(k1), (5.86)

where εµ is the polarization of photon while eµ is the polarization for B. Now we square

this amplitude,

1

4

∑spin

|iM|2 =1

4e2g2gµρgνσ tr

[( γν(/k1 − /p1)γµ

t+γµ(/p1

− /k2)γν

u

)/k1

×( γρ(/k1 − /p1

)γσ

t+γσ(/p1

− /k2)γρ

u

)/k2

]= 8e2g2

[(k1 · p1)(k2 · p1)

t2+

(k1 · p1)(k2 · p1)

u2

+2(k1 · k2)(k1 · k2 − k1 · p1 − k2 · p1)

tu

]= 2e2g2

[u

t+t

u+

2s(s+ t+ u)

tu

]= 2e2g2

[u

t+t

u+

2sM2B

tu

]. (5.87)

Then the cross section can be evaluated as,( dσ

dΩ

)CM

=1

2Ek12Ek2|vk1 − vk2 ||p1|

(2π)24ECM

(14

∑|M|2

)


=e2g2

32π2s

(1− M2

B

s

)[ ut

+t

u+

2sM2B

tu

]. (5.88)

We can also write this differential cross section in terms of squared COM energy s and

scattering angle θ. To do this, we note,

s = E2CM, t = (M2

B − E2CM) sin2 θ

2, u = (M2

B − E2CM) cos2 θ

2. (5.89)

Then we have, ( dσ

dΩ

)CM

=e2g2(1−M2

B/s)

16π2s sin2 θ

[1 + cos2 θ +

4sM2B

(s−M2B)2

], (5.90)

and, ( dσ

d cos θ

)CM

=αg2(1−M2

B/s)

2s sin2 θ

[1 + cos2 θ +

4sM2B

(s−M2B)2

]. (5.91)

(c) The differential cross obtained in (b) diverges when θ → 0 or θ → π. Now let us study

the former case, namely θ → 0.

If we cut of the integral from θ2c ' m2

e/s, then we have,∫θc

( dσ

d cos θ

)CM

sin θdθ ' αg2(1−M2B/s)

2s

[2 +

4sM2B

(s−M2B)2

] ∫ 1−m2e/s dt

1− t2

' αg2(1−M2B/s)

4s

[2 +

4sM2B

(s−M2B)2

]log( s

m2e

)=

αg2

2

1 +M4B/s

2

s−M2B

log( s

m2e

). (5.92)

Now we calculate the following expression,∫ 1

0

dx f(x)σ(e+e− → B)∣∣ECM=(1−x)s

=

∫ 1

0

dx

[α

2π

1 + (1− x)2

xlog( s

m2e

)]πg2δ

(M2

B − (1− x)s)

=αg2

2

1 +M4B/s

2

s−M2B

log( s

m2e

). (5.93)

5.6 The spinor products (3)

This problem generalize the spinor product formalism to the processes involving external

photons.

(a) Firstly we can represent photon’s polarization vectors in terms of spinors of definite

helicity. Let the momentum of the photon be k, and p be a lightlike momentum such that

p · k 6= 0. Then, the polarization vector εµ±(k) of the photon can be taken to be,

εµ+(k) =1√

4p · kuR(k)γµuR(p), εµ−(k) =

1√4p · k

uL(k)γµuL(p), (5.94)


where the spinors uL,R(k) have been introduced in Problems 3.3 and 5.3. Now we use this

choice to calculate the polarization sum,

εµ+(k)εν∗+ (k) + εµ−(k)εν∗− (k)

=1

4p · k

[uR(k)γµuR(p)uR(p)γνuR(k) + uL(k)γµuL(p)uL(p)γνuL(k)

]=

1

4p · ktr[/pγ

ν/kγµ]

= −gµν +pµkν + pνkµ

p · k. (5.95)

When dotted into an amplitude with external photon, the second term of the result vanishes.

This justifies the definitions above for photon’s polarization vectors.

(b) Now we apply the formalism to the process e+e− → 2γ in the massless limit. The

relevant diagrams are similar to those in Figure 5.3, except that one should replace the

label ‘B’ by ‘γ’. To simplify expressions, we introduce the standard shorthand notations as

follows:

p〉 = uR(p), p] = uL(p), 〈p = uL(p), [p = uR(p). (5.96)

Then the spin products become s(p1, p2) = [p1p2] and t(p1, p2) = 〈p1p2〉. Various expres-

sions get simplified with this notation. For example, the Fierz identity (5.37) now reads

[k2γµk1〉[p1γµp2〉 = 2[p1k2]〈k1p2〉. Similarly, we also have 〈k1γ

µk2]〈p1γµp2] = 2〈k1p1〉[p2k2].

Now we write down the expression for tree amplitude of e+Re−L → γRγL. For illustration,

we still keep the original expression as well as all explicit mid-steps. The auxiliary lightlike

momenta used in the polarization vectors are arbitrarily chosen such that the calculation

can be mostly simplified.

iM(e+Re−L → γLγR)

= (−ie)2ε∗−µ(p1)ε∗+ν(p2)uL(k2)

[γν

i

/k1 − /p1

γµ + γµi

/k1 − /p1

γν]uL(k1)

=− ie2 〈k2γµp1][k1γνp2〉4√

(k2 · p1)(k1 · p2)

[ 〈k2γν(/k1 − /p1

)γµk1]

t+〈k2γ

µ(/k1 − /p2)γνk1]

u

]=− ie2 〈k2γµp1][k1γνp2〉

2u

[〈k2γ

νk1]〈k1γµk1]− 〈k2γ

νp1]〈p1γµk1]

t

+〈k2γ

µk1]〈k1γνk1]− 〈k2γ

µp2]〈p2γνk1]

u

]=−2ie2

u

[〈k1k2〉[p1k1]〈k2p2〉[k1k1]− 〈k2p1〉[k1p1]〈k2p2〉[k1p1]

t

+〈k2k2〉[k1p1]〈k1p2〉[k1k1]− 〈k2k2〉[p2p1]〈p2p2〉[k1k1]

u

]= 2ie2 〈k2p1〉[k1p1]〈k2p2〉[k1p1]

tu, (5.97)


where we have used the spin sum identity /p = p]〈p+ p〉[p in the third equality, and also the

Fierz transformations. Note that all spinor products like 〈pp〉 and [pp], or 〈pγµk〉 and [pγµk]

vanish. Square this amplitude, we get∣∣iM(e+Re−L → γLγR)

∣∣2 = 4e4 t

u. (5.98)

In the same way, we calculate other polarized amplitudes,

iM(e+Re−L → γRγL)

=− ie2 [k1γµp1〉〈k2γνp2]

4√

(k1 · p1)(k2 · p2)

[[k2γ

ν(/k1 − /p1)γµk1〉

t+

[k2γµ(/k1 − /p2

)γνk1〉u

]= 2ie2 〈k2p1〉[k1p2]〈k2p2〉[k1p2]

tu. (5.99)

Note that we have used a different set of auxiliary momenta in photons’ polarizations. After

evaluating the rest two nonvanishing amplitudes, we get the squared polarized amplitudes,

as follows: ∣∣M(e+Re−L → γLγR)

∣∣2 =∣∣M(e+

Le−R → γRγL)

∣∣2 = 4e4 t

u, (5.100)∣∣M(e+

Le−R → γLγR)

∣∣2 =∣∣M(e+

Le−R → γLγR)

∣∣2 = 4e4 u

t. (5.101)

Then the differential cross section follows straightforwardly,

dσ

d cos θ=

1

16πs

(1

4

∑spin

|iM|2)

=2πα2

s

( tu

+u

t

), (5.102)

which is in accordance with (5.107) of P&S.


Chapter 6

Radiative Corrections: Introduction

6.1 Rosenbluth formula

In this problem we derive the differential cross section for the electron-proton scattering

in the lab frame, assuming that the scattering energy is much higher than electron’s mass,

and taking account of the form factors of the proton. The result is known as Rosenbluth

formula. The relevant diagram is shown in Figure 6.1.

k1 k2

p1p2

e− p

e− p

Figure 6.1: The electron-proton scattering. The blob denotes form factors that includes

the effect of strong interaction. All initial momenta go inward and all final momenta go

outward.

Let us firstly work out the kinematics. In the lab frame, the momenta can be parame-

terized as

k1 = (E, 0, 0, E), p1 = (E ′, E ′ sin θ, 0, E ′ cos θ), k2 = (M, 0, 0, 0), (6.1)

and p2 can be found by momentum conservation, k1 + k2 = p1 + p2. With the on-shell

condition p22 = M2, we find that

E ′ =ME

M + 2E sin2 θ2

. (6.2)

We also use q = k1 − p1 to denote the momentum transfer and t = q2 its square. Note that

we have set the electron mass to zero.

Now we write down the amplitude M.

iM = (−ie)2U(p2)

[γµF1(q2) +

iσµνqν2M

F2(q2)

]U(k2)

−i

tu(p1)γµu(k1), (6.3)

51

52 Chapter 6. Radiative Corrections: Introduction

where U is the spinor for the proton and u is for the electron, M is the mass of the proton. At

this stage, we convert this expression into a more convenient form by means of the Gordon

identity (see Problem 3.2),

iM = (−ie)2U(p2)

[γµ(F1 + F2)− (p2 + k2)µ

2MF2

]U(k2)

−i

tu(p1)γµu(k1). (6.4)

Now, the squared amplitude with initial spins averaged and final spins summed is,

1

4

∑|M|2 =

e4

4q4tr

[(γµ(F1 + F2)− (p2 + k2)µ

2MF2

)(/k2 +M)

×(γρ(F1 + F2)− (p2 + k2)ρ

2MF2

)(/p2

+M)

]tr[γµ/k1γρ/p1

]=

4e4M2

q4

[(2E2 + 2E ′2 + q2

)(F1 + F2)2

−(

2F1F2 + F 22

(1 + q2

4M2

))((E + E ′)2 + q2

(1− q2

4M2

))]. (6.5)

There are two terms in the square bracket in the last expression. We rewrite the first factor

in the second term as

2F1F2 + F 22

(1 + q2

4M2

)= (F1 + F2)2 − F 2

1 + q2

4M2F22 ,

and combine the (F1 + F2)2 part into the first term, which leads to,

1

4

∑|M|2 =

4e4M2

q4

[q4

2M2 (F1 + F2)2 + 4(F 2

1 −q2

4M2F22

)EE ′ cos2 θ

2

],

where we have used the following two relations which can be easily justified,

E ′ − E = q2

2m, (6.6)

q2 = −4E ′E sin2 θ2. (6.7)

Now we can put the squared amplitude into its final form,

1

4

∑|M|2 =

16e4E2M3

q4(M + 2E sin2 θ

2

)×[(F 2

1 −q2

4M2F22

)cos2 θ

2− q2

2M2(F1 + F2)2 sin2 θ

2

]. (6.8)

On the other hand, we can derive the A+B → 1 + 2 differential cross section in the lab

frame as

dσL =1

2EA2EB|vA − vB|

∫d3p1d3p2

(2π)62E12E2

|M|2(2π)4δ(4)(p1 + p2 − pA − pB). (6.9)

In our case, EA = E, EB = M , E1 = E ′, and |vA − vB| ' 1, thus,

dσL =1

4EM

∫d3p1d3p2

(2π)62E12E2

|M|2(2π)4δ(4)(p1 + p2 − pA − pB)

6.2. Equivalent photon approximation 53

=1

4EM

∫E ′2dE ′d cos θdϕ

(2π)32E ′2E2

|M|2(2π)δ(E ′ + E2(E ′)− E −M

)=

1

4EM

∫E ′2dE ′d cos θdϕ

(2π)22E ′2E2

|M|2[1 +

E ′ − E cos θ

E2(E ′)

]−1

δ

(E ′ − ME

M + 2E sin2 θ2

)=

1

4EM

∫d cos θ

8π|M|2 E ′

M + 2E sin2 θ2

,

where we use the notation E2 = E2(E ′) to emphasize that E2 is a function of E ′. That is,

E2 =√M2 + E2 + E ′2 − 2E ′E cos θ.

Then, ( dσ

d cos θ

)L

=1

32π(M + 2E sin2 θ

2

)2 |M|2. (6.10)

So finally we get the differential cross section, the Rosenbluth formula,( dσ

d cos θ

)L

=πα2

2E2(1 + 2E

Msin2 θ

2

)sin4 θ

2

×[(F 2

1 −q2

4M2F22

)cos2 θ

2− q2

2M2(F1 + F2)2 sin2 θ

2

]. (6.11)

6.2 Equivalent photon approximation

In this problem we study the scattering of a very high energy electron from a target in

the forward scattering limit. The relevant matrix element is,

M = (−ie)u(p′)γµu(p)−igµνq2Mν(q). (6.12)

(a) First, the spinor product in the expression above can be expanded as,

u(p′)γµu(p) = A · qµ +B · qµ + C · εµ1 +D · εµ2 . (6.13)

Now, using the fact that qµu(p′)γµu(p) = 0, we have,

0 = Aq2 +Bq · q ' −4AEE ′ sin2 θ2

+Bq · q ⇒ B ∼ θ2. (6.14)

(b) It is easy to find that

εµ1 = N(0, p′ cos θ − p, 0,−p′ sin θ), εµ2 = (0, 0, 1, 0),

where N = (E2 + E ′2 − 2EE ′ cos θ)−1/2 is the normalization constant. Then, for the right-

handed electron with spinor u+(p) =√

2E(0, 0, 1, 0)T and left-handed electron with u−(p) =√2E(0, 1, 0, 0)T , it is straightforward to show that

u+(p′) =√

2E ′(0, 0, cos θ2, sin θ

2)T , u−(p′) =

√2E ′(− sin θ

2, cos θ

2, 0, 0), (6.15)


and,

u±(p′)γ · ε1u±(p) ' −√EE ′

E + E ′

|E − E ′|θ, (6.16)

u±(p′)γ · ε2u±(p) ' ±i√EE ′θ, (6.17)

u±(p′)γ · ε1u∓(p) = u±(p′)γ · ε2u∓(p) = 0. (6.18)

That is to say, we have,

C± = −√EE ′

E + E ′

|E − E ′|θ, D± = ±i

√EE ′θ. (6.19)

(c) The squared amplitude is given by,

|M±±|2 =e2

(q2)2Mµ(q)Mν(q)

(C±ε

µ1 +D±ε

µ2

)(C∗±ε

ν∗1 +D∗±ε

ν∗2

). (6.20)

Averaging and summing over the initial and final spins of the electron respectively, we get,

1

2

∑|M|2 =

e2

2(q2)2Mµ(q)Mν(q)

[(|C+|2 + |C−|2

)εµ1ε

ν∗1 +

(|D+|2 + |D−|2

)εµ2ε

ν∗2

+(C+D

∗+ + C−D

∗−)εµ1ε

ν∗2 +

(C∗+D+ + C∗−D−

)εµ2ε

ν∗1

]=

e2

(q2)2Mµ(q)Mν(q)EE

′θ2

[( E + E ′

E − E ′)2

εµ1εν∗1 + εµ2ε

ν∗2

]. (6.21)

Then the cross section reads,∫dσ =

1

2E2Mt

∫d3p′

(2π)32E ′d3pt

(2π3)2Et

(1

2

∑|M|2

)(2π)4δ(4)

(∑pi)

=e2

2E2Mt

∫d3p′

(2π)32E ′EE ′θ2

3(q2)2

[( E + E ′

E − E ′)2

+ 1

]×∫

d3pt

(2π3)2Et

|Mµ(q)|2(2π)4δ(4)(∑

pi)

=− 1

2E2Mt

α

2π

∫dx[1 +

( 2− xx

)2] ∫ π

0

dθθ2 sin θ

4(1− cos θ)2

×∫

d3pt

(2π3)2Et

|Mµ(q)|2(2π)4δ(4)(∑

pi). (6.22)

where we have used the trick described in the final project of Part I (radiation of gluon jets)

to separate the contractions of Lorentz indices, and x ≡ (E − E ′)/E. Now let us focus on

the integral over the scattering angle θ in the last expression, which is contributed from the

following factor, ∫ π

0

dθθ2 sin θ

4(1− cos θ)4∼∫

0

dθ

θ. (6.23)

which is logarithmically divergent as θ → 0.

6.3. Exotic contributions to g − 2 55

(d) We reintroduce the mass of the electron into the denominator to cut off the divergence,

namely, let q2 = −2(EE ′ − pp′ cos θ) + 2m2. Then we can expand q2, treating m2 and θ as

small quantities, as,

q2 ' −(1− x)E2θ2 − x2

1− xm2.

Then the polar angle integration near θ = 0 becomes,∫0

dθ θ3

[θ2 +

x2

(1− x)2

m2

E2

]−2

∼ − 1

2log

E2

m2. (6.24)

(e) Combining the results above, the cross section can be expressed as,∫dσ =− 1

2E2Mt

α

2π

∫dx[1 +

( 2− xx

)2] ∫0

dθ θ3

[θ2 +

x2

(1− x)2

m2

E2

]−2

×∫

d3pt

(2π3)2Et

|Mµ(q)|2(2π)4δ(4)(∑

pi)

=1

2E2Mt

α

2π

∫dx

1 + (1− x)2

x2log

E2

m2

×∫

d3pt

(2π3)2Et

|Mµ(q)|2(2π)4δ(4)(∑

pi). (6.25)

6.3 Exotic contributions to g − 2

(a) The 1-loop vertex correction from Higgs boson is,

u(p′)δΓµu(p) =( iλ√

2

)2∫

ddk

(2π)di

(k − p)2 −m2h

u(p′)i

/k + /q −mγµ

i

/k −mu(p)

=iλ2

2

∫ 1

0

dx

∫ 1−x

0

dy

∫ddk′

(2π)d2u(p′)Nµu(p)

(k′2 −∆)3, (6.26)

with

Nµ = (/k + /q +m)γµ(/k +m), (6.27)

k′ = k − xp+ yq, (6.28)

∆ = (1− x)m2 + xm2h − x(1− x)p2 − y(1− y)q2 + 2xyp · q. (6.29)

To put this correction into the following form,

Γµ = γµF1(q) +iσµνqν

2mF2(q), (6.30)

we first rewrite Nµ as,

Nµ = Aγµ +B(p′ + p)µ + C(p′ − p)µ, (6.31)

where term proportional to (p′−p) can be thrown away by Ward identity qµΓµ(q) = 0. This

can be done by gamma matrix calculations, leading to the following result,

Nµ =[(

2d− 1)k′2 + (3 + 2x− x2)m2 + (y − xy − y2)q2

]γµ + (x2 − 1)m(p′ + p)µ. (6.32)


Then, using Gordon identity, we find,

Nµ =[(

2d− 1)k′2 + (x+ 1)2m2 + (y − y2 − xy)q2

]γµ +

iσµνqν2m

· 2m2(1− x2). (6.33)

Comparing this with (6.30), we see that

δF2(q = 0) = 2iλ2m2

∫ 1

0

dx

∫ 1−x

0

dy

∫d4k′

(2π)4

1− x2

(k′2 −∆)3

=λ2

(4π)2

∫ 1

0

dx(1− x)2(1 + x)

(1− x)2 + x(mh/m)2. (6.34)

To carry out the integration over x, we use the approximation that mh m. Then,

δF2(q = 0) ' λ2

(4π)2

∫ 1

0

dx

[1

1 + x(mh/m)2− 1 + x− x2

(mh/m)2

]' λ2

(4π)2(mh/m)2

[log(m2

h/m2)− 7

6

]. (6.35)

(b) According to (a), the limits on λ and mh is given by,

δF2(q = 0) =λ2

(4π)2(mh/m)2

[log(m2

h/m2)− 7

6

]< 1× 10−10. (6.36)

For electron, λ ' 3× 10−6, m ' 0.511MeV, and with mh ∼ 60 GeV, we have δF2(q = 0) ∼10−22 10−10. For realistic case mh ' 125GeV the effect is even smaller. On the other

hand, for muon, we have λ = 6 × 10−4, m ' 106MeV, and with the input mh ' 125GeV,

we have δF2(q = 0) ∼ 10−14. At present the experimentally measured muon’s anomalous

magnetic moment is a bit different from the prediction of Standard Model, and the difference

is of order 10−9, a not decisive but still noteworthy “anomaly”. More can be found in [3].

(c) The 1-loop correction from the axion is given by,

u(p′)δΓµu(p) =( −λ√

2

)2∫

ddk

(2π)di

(k − p)2 −m2h

u(p′)γ5 i

/k + /q −mγµ

i

/k −mγ5u(p)

=− iλ2

2

∫ 1

0

dx

∫ 1−x

0

dy

∫ddk′


(k′2 −∆)3, (6.37)

in which k′ and ∆ are still defined as in (a) except the replacement mh → ma, while Nµ is

now given by,

Nµ = γ5(/k + /q +m)γµ(/k +m)γ5 = −(/k + /q −m)γµ(/k −m). (6.38)

Repeating the same derivation as was done in (a), we get,

Nµ =[−(

2d− 1)k′2 − (1− x− y)yq2 + (1− x)2m2

]γµ − (1− x)2m(p′ + p)2. (6.39)

6.3. Exotic contributions to g − 2 57

Again, using Gordon identity, we get,

δF2(q = 0) =− 2iλ2m2

∫ 1

0

dx

∫ 1−x

0

dy

∫d4k′

(2π)4

(1− x)2

(k′2 −∆)3

=− λ2

(4π)2

∫ 1

0

dx(1− x)3

(1− x)2 + xm2a/m

2

'− λ2

(4π)2

∫ 1

0

dx

[1

1 + xm2a/m

2− 3− 3x+ x2

m2a/m

2

]=− λ2

(4π)2(m2a/m

2)

[log(m2

a/m2)− 11

6

]. (6.40)

For order-of-magnitude estimation, it’s easy to see that λm/ma & 10−5 is excluded.


Chapter 7

Radiative Corrections:

Some Formal Developments

7.1 Optical theorem in φ4 theory

In this problem we check the optical theorem in phi-4 theory to order λ2. Firstly, the

total cross section σtot at this order receives contributions from tree level only. The squared

amplitude is simply λ2. Then its easy to get the total cross section by complementing

kinematic factors. That is,

σtot =λ2

16πs, (7.1)

where s = E2CM and ECM is the COM energy. Then, consider the imaginary part of the

scattering amplitude. The contribution comes from 1-loop diagram in s-channel this time.

Let’s evaluate this amplitude directly,

iM =1

2(−iλ)2

∫ddk

(2π)di

k2 −m2

i

(k − p)2 −m2=

λ2

2

∫ddk′

(2π)d

∫ 1

0

dx1

(k′2 −∆)2

=iλ2

2(4π)2

[2ε− γ + log 4π −

∫ 1

0

dx log(m2 − x(1− x)s

)]. (7.2)

Therefore,

ImM =− λ2

2(4π)2

∫ 1

0

dx Im[

log(m2 − x(1− x)s

)]. (7.3)

The argument in the logarithm is real, thus the imaginary part of the logarithm equals to 0

or −π depending on the argument is positive or negative. (Strictly speaking the imaginary

part is −π but not π due to our iε prescription.) Then we see this logarithm contributes an

constant imaginary part −π, only when

1−√

1− 4m2/s

2< x <

1 +√

1− 4m2/s

2.

Thus we have

ImM =λ2

32π

√1− 4m2/s =

λ2pCM

16πECM

. (7.4)

59

60 Chapter 7. Radiative Corrections: Some Formal Developments

7.2 Alternative regulators in QED

In this problem we compute the first order corrections to Z1 and Z2 in QED, using cut-off

regularization and dimensional regularization respectively. By definition, we have,

Γµ(q = 0) = Z−11 γµ, (7.5)

Z−12 = 1− dΣ

d/p

∣∣∣/p=m

. (7.6)

Let’s begin with dimensional regularization instead of momentum cut-off.

(b) Dimensional Regularization We firstly calculate δF1(0):

u(p′)δΓµ(p, p′)u(p)

=(−ie)2

∫ddk

(2π)d−igρσ

(k − p)2 − µ2u(p′)γσ

i

/k + /q −mγµ

i

/k −mγρu(p)

=− ie2

∫ddk

(2π)du(p′)γρ(/k + /q +m)γµ(/k +m)γρu(p)(

(k − p)2 − µ2)(

(k + q)2 −m2)(k2 −m2

)=− ie2

∫ddk

(2π)d

∫ 1

0

dx

∫ 1−x

0

dy2u(p′)Nµu(p)

(k′2 −∆)3, (7.7)

in which we define

k′ = k − xp+ yq,

∆ = (1− x)m2 + xµ2 − x(1− x)p2 − y(1− y)q2 + 2xyp · q,Nµ = γρ(/k + /q +m)γµ(/k +m)γρ.

The next step is to put Nµ into the needed form. The calculation is basically in parallel

with Peskin’s Sec.6.3. Let me show some details. The first step is to finish the summation

over dummy Lorentz indices. Note that we are using dimensional regularization, thus we

should use Peskin’s Eq.(A.55). The result is:

Nµ = −2/kγµ(/k+/q)+4m(2k+q)µ− (d−2)m2γµ+(4−d)[(/k+/q)γ

µ/k−m(/k+/q)γµ−mγµ/k

].

It’s worth noting here that d will be sent to 4 at the end of the calculation. Thus in the

square bracket in this expression, only the combination /kγµ/k contributes to the final result.

Thus we simply have

Nµ = −2/kγµ(/k + /q) + 4m(2k + q)µ − (d− 2)m2γµ + (4− d)/kγµ/k. (7.8)

Here and following, we are free to drop off terms in Nµ which contribute nothing to final

results. The equal sign should be understood in this way. The next step is to rewrite Nµ in

terms of k′ instead of k:

Nµ = (2− d)/k′γµ/k

′ − 2[x/p− y/q

]γµ[x/p+ (1− y)/q

]+ 4m

[2xp+ (1− 2y)q

]µ − 2m2γµ. (7.9)

7.2. Alternative regulators in QED 61

Terms linear in k′ has been dropped since they integrate to zero. The third step is to put Nµ

into a linear combination of γµ, (p+p′)µ. Terms proportional to (p−p′)µ will be dropped due

to Ward identity. The basic strategy is using substitution qµ = (p′ − p)µ, on shell condition

u(p′)/p′ = u(p′)m and /pu(p) = mu(p). Here we show the detailed steps for the second term

above:

− 2[x/p− y/q

]γµ[x/p+ (1− y)/q

]=− 2

[x2/pγ

µ/p− y(1− y)/qγ

µ/q − xy/qγµ/p+ x(1− y)/pγ

µ/q]

=− 2[x2(2pµ − γµ/p)/p− y(1− y)(2qµ − γµ/q)/q − xy/qγµ/p+ x(1− y)(/p

′ − /q)γµ/q]

=− 2[2x2mpµ − x2m2γµ + y(1− y)q2γµ − 2xymqµ + xmγµ/q − x(1− y)/qγ

µ/q]

=− 2[− x(x+ 2)m2γµ + (x+ y)(1− y)q2γµ + 2x2mpµ + 2xmp′µ

].

Combining this with other terms, and also make the momentum symmetrization (Peskin’s

Eq.(7.87)), which amounts to make the substitution:

/k′γµ/k

′ →(

2d− 1)k′2γµ,

we get

Nµ = (2−d)2

d/k′γµ/k

′+[2(x2 +2x−1)m2−2(x+y)(1−y)q2

]γµ+2x(1−x)m(p′+p)µ. (7.10)

Now we employ Gordon’s identity

u(p′)γµu(p) = u(p′)[ p′µ + pµ

2m+iσµνqν

2m

]u(p),

to put Nµ into a linear combination of γµ and σµν :

Nµ =[ (2−d)2

dk′2 − 2(x2 − 4x+ 1)m2 − 2(x+ y)(1− y)q2

]γµ − 2x(1− x)miσµνqν . (7.11)

Now we have put the vertex Γµ into the following form:

Γµ = γµF1(q) +iσµνqν

2mF2(q). (7.12)

We are interest in δF1(q), which is related to δZ1 by δZ1 = −δF1(q = 0). Finishing

momentum integral:

δF1(0) =− 2ie2

∫ 1

0

dx

∫ 1−x

0

dy

∫ddk′

(2π)d1

(k′2 −∆)3

[ (2−d)2

dk′2 − 2(x2 − 4x+ 1)m2

]=

2e2

(4π)d/2

∫ 1

0

dx

∫ 1−x

0

dy[ (2− d)2Γ(2− d

2)

4∆2−d/2 + (x2 − 4x+ 1)m2 Γ(3− d2

)

∆3−d/2

], (7.13)

and sending d = 4− ε→ 4:

δF1(0) =2e2

(4π)2

∫ 1

0

dx(1− x)[

2ε− γ + log 4π


− log((1− x)2m2 + xµ2

)− 2 +

(x2 − 4x+ 1)m2

(1− x)2m2 + xµ2

], (7.14)

we reach the needed result δZ1 = −δF1(0). Now let us turn to Z2. The correction of first

order is given by δZ2 = (dΣ/d/p)∣∣/p=m

. Therefore we’d better evaluate Σ(/p) using dimensional

regularization.

−iΣ(/p) =(−ie)2

∫ddk

(2π)dγµ

i

/k −mγµ

−i

(p− k)2 − µ2

=− e2

∫ddk

(2π)d(2− d)/k + dm

(k2 −m2)((p− k)2 − µ2

) = −e2

∫ddk′

(2π)d

∫ 1

0

dx(2− d)x/p+ dm

(k′2 −∆)2

=− ie2

(4π)d/2

∫ 1

0

dxΓ(2− d

2)

∆2−d/2

[(2− d)x/p+ dm

], (7.15)

where

k′ = k − xp;∆ = (1− x)m2 + xµ2 − x(1− x)p2.

Then we can compute

dΣ(/p)

d/p=

e2

(4π)d/2

∫ 1

0

dx[ Γ(2− d

2)

∆2−d/2 (2− d)x−(2− d

2)Γ(2− d

2)

∆3−d/2d∆

d/p

((2− d)x/p+ dm

)]=

e2

(4π)d/2

∫ 1

0

dx[ Γ(2− d

2)

∆2−d/2 (2− d)x+Γ(3− d

2)

∆3−d/2 2x(1− x)/p((2− d)x/p+ dm

)].

(7.16)

Then, setting /p = m and d = 4− ε with ε→ 0, we get

dΣ(/p)

d/p

∣∣∣/p=m

=−2e2

(4π)2

∫ 1

0

dx x[

2ε

+ γ + log 4π

− log((1− x)2m2 + xµ2

)− 1− 2(1− x)(2− x)m2

(1− x)2m2 + xµ2

](7.17)

Now it’s still not easy to see δZ1 = δZ2 immediately. To make this transparent, we need

some more steps. Let’s focus on logarithm term:

−∫ 1

0

dx(1− x) log((1− x)2m2 + xµ2

)= −

∫ 1

0

dx(1− 2x+ x) log((1− x)2m2 + xµ2

)=

∫ 1

0

dx[(1− x)− (1− x)(1− x2)m2

(1− x2)m2 + xµ2− x log

((1− x)2m2 + xµ2

)](7.18)

Combining this with other terms, and also using the fact∫xdx =

∫(1− x)dx, we get

δF1(0) =2e2

(4π)2

∫ 1

0

dxx[

2ε− γ + log 4π

− log((1− x)2m2 + xµ2

)− 1− 2(1− x)(2− x)m2

(1− x)2m2 + xµ2

]. (7.19)

Now it’s clear that δZ1 = δZ2. Thus Z1 = Z2 keeps unaffected at this order when dimensional

regularization is used.

7.3. Radiative corrections in QED with Yukawa interaction 63

(a) Momentum Cut-off Now we repeat the calculation above using momentum cut-off.

Now we can directly borrow some results above. All we need to do is setting d → 4 and

adding a UV momentum cut-off Λ, as well as the following integral formulae:∫ Λ d4k

(2π)4

1

(k2 −∆)2=

i

16π2

[log(1 + Λ2

∆

)− Λ2

Λ2+∆

],∫ Λ d4k

(2π)4

k2

(k2 −∆)3=

i

16π2

[log(1 + Λ2

∆

)+ ∆(4Λ2+3∆)

2(Λ2+∆)2− 3

2

],∫ Λ d4k

(2π)4

1

(k2 −∆)3= − i

32π2

Λ4

∆(Λ2 + ∆)2.

We begin with (7.13):

δF1(0) =− 2ie2

∫ 1

0

dx

∫ 1−x

0

dy

∫ Λ d4k′

(2π)4

1

(k′2 −∆)3

[k′2 − 2(x2 − 4x+ 1)m2

]=

e2

8π2

∫ 1

0

dx (1− x)

[log(

1 +Λ2

∆

)+

(x2 − 4x+ 1)m2

∆− 3

2

], (7.20)

In the same way, we get

−iΣ(/p) = 2e2

∫ Λ d4k′

(2π)4

∫ 1

0

dxx/p− 2m

(k′2 −∆)2, (7.21)

and

dΣ(/p)

d/p

∣∣∣/p=m

=−e2

8π2

∫ 1

0

dx

[x log

(1 +

Λ2

∆

)− x+

2x(1− x)(x− 2)m2

∆

]. (7.22)

This shows that δZ1 6= δZ2 with cut-off regularization.

7.3 Radiative corrections in QED with Yukawa inter-

action

(a) Let us calculate the first order corrections to Z1 and Z2, as was done in Problem 7.2.

Firstly, we calculate δΓµ, which is similar to the corresponding QED correction:

u(p′)δΓµ(p, p′)u(p)

=(−iλ2/√

2)2

∫ddk

(2π)di

(k − p)2 −m2φ

u(p′)i

/k + /q −mγµ

i

/k −mu(p)

=iλ2

2

∫ 1

0

dx

∫ 1−x

0

dy

∫ddk


(k′2 −∆)3, (7.23)

where

k′ = k − xp+ yq,


∆ = (1− x)m2 + xm2φ − x(1− x)p2 − y(1− y)q2 + 2xyp · q,

Nµ = (/k + /q +m)γµ(/k +m).

Then we put this correction into the following form, in parallel with steps of Problem 7.2.

That is: (1) replace k by k′ in Nµ:

Nµ = /k′γµ/k

′+(x/p+ (1− y)/q +m

)γµ(x/p− y/q +m);

(2) rewrite the numerator Nµ by gamma matrix relations and equations of motion, as

Nµ =[(

2d− 1)k′2 + (3 + 2x− x2)m2 + (y − xy − y2)q2

]γµ + (x2 − 1)m(p′ + p)µ;

(3) use Gordon identity to further transform Nµ into:

Nµ =[(

2d− 1)k′2 + (x+ 1)2m2 + (y − y2 − xy)q2

]γµ +

iσµν

2m· 2m2(1− x2).

Then, we can read off δF1 from the coefficient of γµ, as:

δF1(0) = iλ2

∫ 1

0

dx

∫ 1−x

0

dy

∫ddk

(2π)d1

(k′2 −∆)3

[(2d− 1)k′2 + (x+ 1)2m2

]=

λ2

2(4π)2

∫ 1

0

dx (1− x)

[2ε− γ + log 4π

− log((1− x)2m2 + xm2

φ

)− 1 +

(x+ 1)2m2

(1− x)2m2 + xm2φ

]. (7.24)

Using the trick identity (7.18) again, we finally get

δF1(0) =λ2

2(4π)2

∫ 1

0

dx x

[2ε− γ + log 4π

− log((1− x)2m2 + xm2

φ

)+

2(1− x2)m2

(1− x)2m2 + xm2φ

]. (7.25)

Now we calculate Σ(/p).

−iΣ(/p) = (−iλ/√

2)2

∫ddk

(2π)di

/k −mi

(p− k)2 − µ2=

λ2

2

∫ddk′

(2π)d

∫ 1

0

dxx/p+m

(k′2 −∆)2

=iλ2

2(4π)2

∫ 1

0

dx[

2ε− γ + log 4π − log ∆

](x/p+m), (7.26)

where k′ = k − xp and ∆ = (1− x)m2 + xµ2 − x(1− x)p2. Then we have

dΣ(/p)

d/p

∣∣∣/p=m

=−λ2

2(4π)2

∫ 1

0

dx x[

2ε− γ + log 4π

− log((1− x)2m2 + xm2

φ

)+

2(1− x2)m2

(1− x)2m2 + xm2φ

]. (7.27)

Thus we have proved that δZ1 = δZ2 holds for 1-loop scalar corrections.

7.3. Radiative corrections in QED with Yukawa interaction 65

(b) Now consider the 1-loop corrections to Yukawa vertex. We focus on the divergent part

only. The equalities below should be understood to be hold up to a finite part. Then, for

vertex correction from photon, we have

δΓ(p, p′)∣∣photon

=(−ie)2

∫ddk

(2π)d−i

(k − p)2 − µ2γµ

i

/k + /q −mi

/k −mγµ

=− ie2

∫ 1

0

dx

∫ 1−x

0

dy

∫ddk

(2π)d2dk′2

(k′2 −∆)3

=d2e2

2(4π)d/2

∫ 1

0

dx

∫ 1−x

0

dyΓ(2− d

2)

∆2−d/2 =4e2

(4π)2

2

ε(7.28)

In the same way,

δΓ(p, p′)∣∣scalar

=( −iλ√

2

)2∫

ddk

(2π)di

(k − p)2 −m2φ

i

/k + /q −mi

/k −m

=iλ2

2

∫ 1

0

dx

∫ 1−x

0

dy

∫ddk

(2π)d2k′2

(k′2 −∆)3=−λ2

2(4π)2

2

ε(7.29)

On the other hand, the 1-loop corrections for electron’s self-energy also come from two

parts: one is the photon correction, which has been evaluated in Problem 7.1,

−iΣ(/p)∣∣photon

= − ie2

(4π)d/2

∫ 1

0

dxΓ(2− d

2)

∆2−d/2

[(2− d)x/p+ dm

]=

ie2(/p− 4m)

(4π)2

2

ε, (7.30)

and the other is the scalar correction:

−iΣ(/p)∣∣scalar

=( −iλ√

2

)2∫

ddk

(2π)di

/k −mi

(p− k)2 −m2φ

=iλ2(/p+ 2m)

4(4π)2

2

ε, (7.31)

To sum up, we have got the total vertex correction:

δΓ(p, p′) = δΓ(p, p′)∣∣photon

+ δΓ(p, p′)∣∣scalar

=4e2 − λ2/2

(4π)2

2

ε, (7.32)

and also:

dΣ(/p)

d/p

∣∣∣/p=m

=d[Σ(/p)photon + Σ(/p)scalar

]d/p

∣∣∣/p=m

= − e2 + λ2/4

(4π)2

2

ε. (7.33)


Final Project I

Radiation of Gluon Jets

In this final project we do a basic exercise about cancellation of infrared divergence. An

excellent pedagogical treatment of infrared divergence can be found in [2]

(a) First we calculate the 1-loop vertex correction to M(e+e− → qq) from virtual gluon.

The amplitude is given by

iδ1M = Qf (−ie)2(−ig)2u(p1)

[ ∫ddk

(2π)dγν

i

/kγµ

i

/k − /qγν

−i

(k − p1)2 − µ2

]v(p2)

−i

q2v(k2)γµu(k1).

(7.34)

Now we simplify the loop integral in the standard way, as was done in Problem 7.2. The

result is

iδ1M = − ig2

[ ∫ddk

(2π)d

∫ 1

0

dx

∫ 1−x

0

dy2(

(2−d)2

dk′2 − 2(1− x)(x+ y)q2

)(k′2 −∆)3

]iM0

=2g2

(4π)d/2

∫ 1

0

dx

∫ 1−x

0

dy[ (2− d)2

4∆2−d/2 Γ(2− d2

) +(1− x)(x+ y)q2

∆3−d/2 Γ(3− d2

)]iM0

=2g2

(4π)2

∫ 1

0

dx

∫ 1−x

0

dy[

2ε− γ + log 4π − log ∆− 2 +

(1− x)(x+ y)q2

∆

]iM0,

(7.35)

where

iM0 = Qf (−ie)2u(p1)γµv(p2)1

q2v(k2)γµu(k1) (7.36)

is the tree amplitude, and

k′ = k − xq − yp1, ∆ = −x(1− x− y)q2 − y(1− y)p21 + yµ2. (7.37)

With the external legs amputated, the result is,

iδ1M =2g2

(4π)2

∫ 1

0

dx

∫ 1−x

0

dy[

log( yµ2

yµ2 − x(1− x− y)q2

)+

(1− x)(x+ y)q2

yµ2 − x(1− x− y)q2

]iM0,

(7.38)

67

68 Final Project I. Radiation of Gluon Jets

Then the cross section is given by

σ(e+e− → qq) =4πα2

3s· 3|F1(q2 = s)|2, (7.39)

with

F1(q2 = s) = Q2f +

Q2fαg

2π

∫ 1

0

dx

∫ 1−x

0

dy[

log( yµ2

yµ2 − x(1− x− y)s

)+

(1− x)(x+ y)s

yµ2 − x(1− x− y)s

]. (7.40)

We will carry out the Feynman integration in (e).

(b) Now we simplify the 3-body phase space integral∫dΠ3 =

∫d3k1d3k2d3k3

(2π)92E12E22E3

δ(4)(q − k1 − k2 − k3) (7.41)

in the center of mass frame. It is convenient to introduce a new set of variables xi = 2ki ·q/q2,

(i = 1, 2, 3). In COM frame, xi = 2Ei/Eq Then one can show that all Lorentz scalars

involving final states only can be represented in terms of xi and particles masses. In fact,

we only need to check (k1 + k2)2, (k2 + k3)2 and (k3 + k1)2. From instance,

(k1 + k2)2 = (q − k3)2 = q2 +m23 − 2q · k3 = s(1− x3) +m2

3. (7.42)

Similarly,

(k2 + k3)2 = s(1− x1) +m21, (k3 + k1)2 = s(1− x2) +m2

2. (7.43)

To simply the phase integral, we first integrate out k3 with spatial delta function that

restricts k3 = k1 + k2:∫dΠ3 =

∫d3k1d3k2

(2π)62E12E22E3

(2π)δ(Eq − E1 − E2 − E3). (7.44)

The integral measure can be rewritten as

d3k1d3k2 = k21k

22dk1dk2dΩ1dΩ12, (7.45)

where dΩ1 is the spherical integral measure associated with d3k1, and dΩ12 is the spherical

integral of relative angles between k1 and k2. The former spherical integral can be directly

carried out and results in a factor 4π. To finish the integral with dΩ12 = d cos θ12dϕ12, we

make use of the remaining delta function, which can be rewritten as

δ(Eq − E1 − E2 − E3) =E3

k1k2

δ(

cos θ12 −E2

3 − k21 − k2

2 − µ2

2k1k2

), (7.46)


by means of E3 =√k2

1 + k22 + 2k1k2 cos θ + µ2. Thus∫

dΩ1dΩ12 δ(Eq − E1 − E2 − E3) =8π2E3

k1k2

.

Now using k1dk1 = E1dE1 and k2dk2 = E2dE2, we have∫dΠ3 =

∫dk1dk2 k

21k

22

8(2π)5E1E2E3

8π2E3

k1k2

=1

32π3

∫dE1dE2 =

s

128π3

∫dx1dx2. (7.47)

To determine the integral region for m1 = m2 = 0 and m3 = µ, we note that there are two

extremal cases: k1 and k2 are parallel or antiparallel. In the former case, we have

Eq = E1 + E2 + E3 = E1 + E2 +√

(E1 + E2)2 + µ2, (7.48)

which yields

2Eq(E1 + E2) = E2q − µ2, (7.49)

while in the latter case,

Eq = E1 + E2 +√

(E1 − E2)2 + µ2, (7.50)

which gives

(Eq − 2E1)(Eq − 2E2) = µ2. (7.51)

These two boundary cases can be represented by xi variables as

x1 + x2 = 1− µ2

s; (7.52)

(1− x1)(1− x2) =µ2

s. (7.53)

The integral thus goes over the region bounded by these two curves.

(c) Now we calculate the differential cross section for the process e+e− → qgg to lowest

order in α and αg. First, the amplitude is

iM = Qf (−ie)2(−ig)ε∗ν(k3)u(k1)

[γν

i

/k1 + /k3

γµ − γµ i

/k2 + /k3

γν]v(k2)

−i

q2v(p2)γµu(p1).

(7.54)

Then, the squared amplitude is

1

4

∑|iM|2 =

Q2fg

2e4

4s2(−gνσ) tr (γµ/p1

γρ/p2)

× tr

[(γν

1

/k1 + /k3

γµ − γµ 1

/k2 + /k3

γν)/k2

(γρ

1

/k1 + /k3

γσ − γσ 1

/k2 + /k3

γρ)/k1

]=

4Q2fg

2e4

3s2

(8p1 · p2

)[ 4(k1 · k2)(k1 · k2 + q · k3)

(k1 + k3)2(k2 + k3)2


+( 1

(k1 + k3)4+

1

(k2 + k3)4

)(2(k1 · k3)(k2 · k3)− µ2(k1 · k2)

)]. (7.55)

We have used the trick described in Peskin’s book (P261) when getting through the last

equal sign. Now rewrite the quantities of final-state kinematics in terms of xi, and set µ→ 0,

we obtain

1

4

∑|iM|2 =

2Q2fg

2e4

3s2

(8p1 · p2

)[ 2(1− x3)

(1− x1)(1− x2)+

1− x1

1− x2

+1− x2

1− x1

]=

8Q2fg

2e4

3s

x21 + x2

2

(1− x1)(1− x2). (7.56)

Thus the differential cross section, with 3 colors counted, reads

dσ

dx1dx2

∣∣∣COM

=1

2Ep12Ep2|vp1 − vp2|s

128π3

(14

∑|M|2

)=

4πα2

3s· 3Q2

f ·αg2π

x21 + x2

2

(1− x1)(1− x2), (7.57)

where we have used the fact that the initial electron and positron are massless, which implies

that 2Ep1 = 2Ep2 =√s and |vp1 − vp2 | = 2 in COM frame.

(d) Now we reevaluate the averaged squared amplitude, with µ kept nonzero in (7.55).

The result is

1

4

∑|iM|2 =

8Q2fg

2e4

3sF (x1, x2, µ

2/s), (7.58)

where

F(x1, x2,

µ2

s

)=

2(x1 + x2 − 1 + µ2

s)(1 + µ2

s)

(1− x1)(1− x2)

+[ 1

(1− x1)2+

1

(1− x2)2

]((1− x1)(1− x2)− µ2

s

). (7.59)

The cross section, then, can be got by integrating over dx1dx2:

σ(e+e− → qqg) =1

2Ep12Ep2|vp1 − vp2 |s

128π3

∫dx1dx2

(14

∑|M|2

)=

4πα2

3s· 3Q2

f ·αg2π

∫ 1−µ2

s

0

dx1

∫ 1− ts(1−x1)

1−x1−µ2

s

dx2 F(x1, x2,

µ2

s

)=

4πα2

3s· 3Q2

f ·αg2π

[log2 µ

2

s+ 3 log

µ2

s+ 5− 1

3π2 +O(µ2)

]. (7.60)

(e) It is straightforward to finish the integration over Feynman parameters in (a), yielding

F1(q2 = s) = Q2f −

Q2fαg

4π

[log2 µ

2

s+ 3 log

µ2

s+ 7

2− 1

3π2 − iπ

(2 log

µ2

s+ 7)

+O(µ2)

].

(7.61)

Then the cross section, to the order of αg, is given by

σ(e+e− → qq) =4πα2

3s· 3Q2

f

1− αg

2π

[log2 µ

2

s+ 3 log

µ2

s+ 7

2− 1

3π2]

+O(µ2)

. (7.62)


(f) Combining the results in (d) and (e), we reach the final result:

σ(e+e− → qq + qqg) =4πα2

3s· 3Q2

f

[1 +

3αg4π

]. (7.63)

It is worth noting that all divergent terms as µ→ 0 cancel out in this expression.


Chapter 9

Functional Methods

9.1 Scalar QED

The Lagrangian for scalar QED reads,

L = − 14FµνF

µν + (Dµφ)†(Dµφ)−m2φ†φ, (9.1)

with

Fµν = ∂µAν − ∂νAµ, Dµφ = (∂µ + ieAµ)φ. (9.2)

(a) Expanding the covariant derivative, it’s easy to find the corresponding Feynman’s

rules:

(Aµ-Aν-φ†-φ interaction) = 2ie2ηµν ,

(Aµ-φ†(p1)-φ(p2) interaction) = − ie(p1 − p2)µ,

with all momenta pointing inwards.

The propagators are standard. We will work in the Feynman gauge and set ξ = 1, then

the propagator for photon is simply−iηµνp2 + iε

,

and the propagator for scalar isi

p2 −m2 + iε.

(b) Now we calculate the spin-averaged differential cross section for the process e+e− →φ∗φ. The scattering amplitude is given by

iM = (−ie)2v(k2)γµu(k1)−i

s(p1 − p2)µ. (9.3)

Then the spin-averaged and squared amplitude is

1

4

∑spins

|M|2 =e4

4s2tr[(/p1− /p2

)/k1(/p1− /p2

)/k2

]73

74 Chapter 9. Functional Methods

=e4

4s2

[8(k1 · p1 − k1 · p2)(k2 · p1 − k2 · p2)− 4(k1 · k2)(p1 − p2)2

]. (9.4)

We may parameterize the momenta as

k1 = (E, 0, 0, E), p1 = (E, p sin θ, 0, p cos θ),

k2 = (E, 0, 0,−E), p2 = (E,−p sin θ, 0,−p cos θ),

with p =√E2 −m2. Then we have

1

4

∑spins

|M|2 =e4p2

2E2sin2 θ. (9.5)

Thus the differential cross section is:( dσ

dΩ

)CM

=1

2(2E)2

p

8(2π)2E

(14

∑|M|2

)=

α2

8s

(1− m2

E2

)3/2

sin2 θ. (9.6)

(c)

δΠµν = 2ie2ηµν

∫ddk

(2π)di

k2 −m2− (−ie)2

∫ddk

(2π)d(p− 2k)µ(p− 2k)ν

(k2 −m2)((p− k)2 −m2)

=− e2

∫ddk

(2π)d2ηµν

((p− k)2 −m2

)− (p− 2k)µ(p− 2k)ν

(k2 −m2)((p− k)2 −m2

)=− e2

∫ddk′

(2π)d

∫ 1

0

dx2ηµν

(k′2 + (1− x)2p2 −m2

)+ (1− 2x)2pµpν + 4k′µk′ν

(k′2 −∆)2

. =− e2

∫ddk′

(2π)d

∫ 1

0

dx2ηµνk

′2(1− 2d

) + 2ηµν((1− x)2p2 −m2

)− (1− 2x)2pµpν

(k′2 −∆)2

=−ie2

(4π)d/2

∫ 1

0

dx[ (1− d

2)Γ(1− d

2)2ηµν

∆2−d/2

+Γ(2− d

2)

∆2−d/2

(2ηµν

((1− x)2p2 −m2

)− (1− 2x)2pµpν

)]=−ie2

(4π)d/2

∫ 1

0

dxΓ(2− d

2)

∆2−d/2

[2((1− x)2 − x(1− x)

)p2ηµν − (1− 2x)2pµpν

]. (9.7)

We can symmetrize the integrand as (1− x)2 → 12

((1− x)2 + x2

), then we get

δΠµν =−ie2

(4π)d/2

∫ 1

0

dxΓ(2− d

2)

∆2−d/2 (1− 2x)2(p2ηµν − pµpν). (9.8)

9.2 Statistical field theory

In this problem we study the path integral formulation in statistical mechanics. The

theory can be described by the partition function:

Z = tr e−βH , (9.9)

9.2. Statistical field theory 75

where H is the Hamiltonian of the system. It is a function of the generalized coordinates

q and the corresponding conjugate momentum p. In this problem, we simply assume the

Hamiltonian has the following form:

H =p2

2m+ V (q). (9.10)

We assume the dimension of the configuration space is d, then both q and p have d compo-

nents. Then we assume the eigenstates of both q and p form a complete orthonormal basis

of the Hilbert space:

1 =

∫ddq |q〉〈q|; 1 =

∫ddp

(2π)d|p〉〈p|. (9.11)

Then the partition function can be written as

Z = tr e−βH =

∫ddq 〈q|e−βH |q〉. (9.12)

(a) Now we derive a path integral expression for the partition function. Following the

same way of deriving path integral in a quantum field theory, we separate the quantity e−βH

into N factors:

e−βH = e−εH · · · e−εH , (N factors),

then inserting a complete basis between each pair of adjacent factors, as

e−βH =

∫ddq1 · · · ddqN−1〈q|e−εH |qN−1〉〈qN−1|e−εH |qN−2〉 · · · 〈q1|e−εH |q〉.

Now we focus on one factors:

〈qi+1|e−εH |qi〉 = 〈qi+1|e−ε(

12m

p2+V (q))|qi〉 = e−εV (qi)〈qi+1|e−

ε2m

p2|qi〉,

and

〈qi+1|e−ε

2mp2 |qi〉 =

∫ddpi+1ddpi(2π)d(2π)d

〈qi+1|pi+1〉〈pi+1|e−εp2/2m|pi〉〈pi|qi〉

=

∫ddp

(2π)deip(qi+1−qi)e−εp

2/2m =[m

2πε

]d/2e−m(qi+1−qi)2/2ε.

Inserting all this into the partition function, we get:

Z =[ m

2πε

]Nd/2 N∏i=0

∫ddqi exp

[− m(qi+1 − qi)2

2ε− εV (qi)

], (9.13)

with qN+1 = q0.

Now let N →∞, then we have

Z =

∫Dq exp

[− β

∮dτLE(τ)

], (9.14)


where the integral measure is defined by

Dq = limN→∞

[ m

2πε(N)

]Nd/2 N∏i=0

ddqi, (9.15)

and LE(τ) is a Lagrangian in Euclidean form:

LE(τ) =m

2

( dq

dτ

)2

+ V (q(τ)). (9.16)

Note that the periodic integral on τ comes from the trace in the partition function.

(b) Now we study an explicit example, a simple harmonic oscillator, which can by defined

by the Lagrangian

LE = 12q2 + 1

2ω2q2. (9.17)

Our task is to complete the path integral to find a expression for the partition function of

harmonic oscillator. This can be easily done by a Fourier transformation of the coordinates

q(τ) with respect to τ . Since the “time” direction is periodic, the Fourier spectrum of q is

discrete. That is,

q(τ) = β−1/2∑n

e2πinτ/βqn, (9.18)

Then we have:∫dτ LE(τ) =

∫dτ

1

2β

∑m,n

[( 2πi

β

)2

mn+ ω2]qmqne

2πi(m+n)τ/β

=1

2

∑m,n

[( 2πi

β

)2

mn+ ω2]qmqnδm,−n =

1

2

∑n∈Z

[( 2π

β

)2

n2 + ω2]qnq−n

=1

2

∑n∈Z

[( 2π

β

)2

n2 + ω2]|qn|2. (9.19)

Then the path integral can be written as,

Z = C

∫dq0 e

−βω2q20

∫ ∏n>0

dReqndImqn exp[− β

2

( 4π2n2

β2+ ω2

)|qn|2

]=

C

ω

∏n>0

[ 4π2n2

β2+ ω2

]−1

=C

ω

∏n>0

[1 +

1

(πn)2

( βω2

)2]−1

= C sinh−1(βω/2) = C∑n≥0

exp[− βω(n+ 1

2)]. (9.20)

(c) From now on we will consider the statistics of fields. We study the statistical properties

of boson system, fermion system, and photon system.

For a scalar field, the Lagrangian is given by,

LE(τ) =

∫d3x

1

2

[φ2(τ,x) +

(∇φ(τ,x)

)2+m2φ2(τ,x)

]. (9.21)

9.2. Statistical field theory 77

Following the method we used to deal with the simple harmonic oscillator, here we decom-

pose the scalar field φ(τ,x) into eigenmodes in momentum space:

φ(τ,x) = β−1/2∑n

e2πinτ/β

∫d3k

(2π)3eik·xφn,k. (9.22)

Then the Lagrangian can also be rewritten in terms of modes, as,∫dτ LE(τ) =

∫dτd3x

∑n,n′

∫d3kd3k′

(2π)6

1

2β

[( 2πi

β

)2

n′n− k′ · k +m2

]× φn,kφn′,k′ei2π(n′+n)τ/β+i(k+k′)·x

=1

2

∑n

∫d3k

(2π)3

[( 2π

β

)2

n2 + k2 +m2

]|φn,k|2

=

∫d3k

(2π)3

[1

2ω2k|φ0,k|2 +

∑n>0

(( 2π

β

)2

n2 + ω2k

)|φn,k|2

], (9.23)

where ω2k = k2 +m2. Then the partition function, as a path integral over the field configu-

rations can be represented by

Z = C

∫ ∏n>0,k

Reφn,kImφn,k exp

[− β

(( 2π

β

)2

n2 + ω2k

)|φn,k|2

]. (9.24)

By the calculation similar to that in (b), we get

Z = C∏k

[ωk

∏n>0

(4π2n2

β2+ ω2

k

)]−1

= C∏k

exp[− βωk

(n+ 1

2

)]. (9.25)

This product gives the meaning to the formal expression[

det(−∂2 +m2)]−1/2

with proper

regularization.

(d) Then consider the fermionic oscillator. The action is given by,

S =

∫dτ LE(τ) =

∫dτ(ψ(τ)ψ(τ) + ωψ(τ)ψ(τ)

). (9.26)

The antiperiodic boundary condition ψ(τ + β) = −ψ(τ) is crucial to expanding the fermion

into modes:

ψ(τ) = β−1/2∑

n∈Z+1/2

e2πiτ/βψn. (9.27)

Then the partition function can be evaluated to be

Z =

∫ ∏n

dψndψn

[− β

∑n∈Z+1/2

ψn

(2πin

β+ ω

)ψn

]

= C(β)∏

n∈Z+1/2

( 2πin

β+ ω

)= C(β)

∞∏n=0

( 4π2(n+ 12

)2

β2+ ω2

)= C(β) cosh

(12βω)

= C(β)(eβω/2 + e−βω/2

), (9.28)

with the form of a two-level system, as expected.


(e) Finally we consider the system of photons. The partition function is given by

Z =

∫DAµDbDc exp

[ ∫dτd3x

(− 1

2Aµ∂

2Aµ − b∂2c)]

= C(β)[

det(−∂)]4·(−1/2) · det(−∂2), (9.29)

where the first determinant comes from the integral over the vector field Aµ while the second

one comes from the integral over the ghost fields. Therefore,

Z = C(β)[

det(−∂2)]2·(−1/2)

, (9.30)

which shows the contributions from the two physical polarizations of a photon. Here we see

the effect of the ghost fields of eliminating the additional two unphysical polarizations of a

vector field.

Chapter 10

Systematics of Renormalization

10.1 One-Loop structure of QED

(a) In this problem we show that any photon n-point amplitude with n an odd number

vanishes.

Now we evaluate explicitly the one-point photon amplitude and three-point photon am-

plitude at 1-loop level to check Furry’s theorem. The one-point amplitude at 1-loop level is

simply given by,

iΓ(1) = (−ie)

∫ddk

(2π)d−i tr [γµ(/k +m)]

k2 −m2= 0, (10.1)

and the three-point amplitude consists of two diagrams,

iΓ(3) = (−ie)3

∫ddk

(2π)d(−1)

tr[γµ

i

/k −mγν

i

/k + /p1−m

γλi

/k + /p1+ /p2

−m

]+ tr

[γµ

i

/k + /p1+ /p2

−mγλ

i

/k + /p1−m

γνi

/k −m

]. (10.2)

(b) Next we will show that the potential logarithmic divergences in photon four-point

diagrams cancel with each other. Since the divergence in this case does not depend on

external momenta, we will set all external momenta to be zero for simplicity. For the same

reason we will also set the fermion’s mass to be zero. Then the six diagrams contributing

the four-point amplitude can be evaluated as,

(Divergent part of iΓµνρσ)

=

∫ddk

(2π)d−1

(k2)4

[tr [γµ/kγν/kγρ/kγσ/k] + tr [γµ/kγν/kγσ/kγρ/k] + tr [γµ/kγρ/kγν/kγσ/k]

+ tr [γµ/kγρ/kγσ/kγν/k] + tr [γµ/kγσ/kγν/kγρ/k] + tr [γµ/kγσ/kγρ/kγν/k]]. (10.3)

Now let’s focus on the first trace, which can be worked out explicitly, to be

tr [γµ/kγν/kγρ/kγσ/k] = 32kµkνkρkσ − 8k2(kµkνgρσ + kρkσgµν + kµkσgνρ + kνkρgµσ

)79

80 Chapter 10. Systematics of Renormalization

+ 4(k2)2(gµνgρσ − gµρgνσ + gµσgνρ). (10.4)

Then, we symmetrize the momentum factors according to kµkν → k2gµν/4 and kµkνkρkσ →(k2)2(gµνgρσ + gµρgνσ + gµσgνρ)/24. (Since the divergence can be at most logarithmic, so it

is safe to set spacetime dimension d = 4 at this stage.) Then the first trace term reduces to,

tr [γµ/kγν/kγρ/kγσ/k]⇒ 4

3(k2)2(gµνgρσ − 2gµρgνσ + gµσgνρ). (10.5)

The other five terms can be easily got by permuting indices. Then it is straightforward to

see that the six terms sum to zero.

10.2 Renormalization of Yukawa theory

In this problem we study the pseudoscalar Yukawa Lagrangian,

L = 12

(∂µφ)2 − 12m2φ2 + ψ(i/∂ −M)ψ − igψγ5ψφ , (10.6)

where φ is a real scalar and ψ is a Dirac Fermion.

(a) Let’s figure out how the superficial degree of divergence D depends on the number of

external lines. From power counting, it’s easy to see that D can be represented by

D = 4L− Pf − 2Ps , (10.7)

where L is the no. of loops, Pf is the no. of internal fermion lines, and Ps is the no. of

internal scalar lines. We also note the following simple relations:

L = Pf + Ps − V + 1 ,

2V = 2Pf +Nf ,

V = 2Ps +Ns .

Then we can deduce

D = 4L− Pf − 2Ps = 4(Pf + Ps − V + 1)− Pf − 2Ps = 4− 32Nf −Ns . (10.8)

Guided by this result, we can find all divergent amplitudes as follows.

D = 2 D = 1

D = 0 D = 0

We note that we have ignored the vacuum diagram, which simply contributes an infinitely

large constant, the potentially divergent diagrams with odd number of external scalars are

also ignored, since they actually vanish. This result shows that the original theory cannot

be renormalized unless we including a new φ4 interaction, as

δL = − λ4!φ4 . (10.9)

10.2. Renormalization of Yukawa theory 81

(b) Now let us evaluate the divergent parts of all 1-loop diagrams of Yukawa theory. First

we consider the two point function of scalar. The one-loop contribution to this amplitude

is shown as follows.

+ +

The d = 4 pole of first two loop diagrams can be determined as

=−iλ

2

∫ddk

(2π)di

k2 −m2∼ iλm2

(4π)2

1

ε. (10.10)

= −(−ig)2

∫ddk

(2π)dtr[ i

/k −Mγ5 i

(/k − /p)−Mγ5]∼ 4ig2(p2 − 2M2)

(4π)2

1

ε.

(10.11)

From this we find the divergent part of the counterterm to be

δm ∼(λm2 − 8g2M2)

(4π)2

1

ε, δφ =

−4g2

(4π)2

1

ε. (10.12)

Then we come to the two point function of fermion, the 1-loop correction of which is given

by the following two diagrams.

+

From the pole of the loop diagram

= g2

∫ddk

(2π)dγ5 i

/k −Mγ5 i

(k − p)2 −m2∼

ig2(/p− 2M)

(4π)2

1

ε, (10.13)

we find the following counterterms:

δM ∼−2g2M

(4π)2

1

ε, δψ ∼

−g2

(4π)2

1

ε. (10.14)

The following two diagrams contribute to 1-loop corrections to Yukawa coupling and φ4

coupling, respectively.


Since the divergent part of diagram is independent of external momenta, we can set all these

momenta to be zero. Then the loop diagram is

= g3

∫ddk

(2π)dγ5 i

/k −Mγ5 i

/k −Mγ5 i

k2 −m2∼ − g3γ5

(4π)2

2

ε(10.15)

=(−iλ)2

2

∫ddk

(2π)d

( i

k2 −m2

)2

∼ iλ2

(4π)2

1

ε. (10.16)

= (−1)g4

∫ddk

(2π)dtr[(γ5 i

/k −M

)4]∼ − 8ig4

(4π)2

1

ε. (10.17)

Note that there are 3 permutations for the first diagram and 6 permutations for the second

diagram. Then we can determine the divergent part of counterterm to be

δg ∼2g3

(4π)2

1

ε, δλ ∼

3λ2 − 48g4

(4π)2

1

ε. (10.18)

10.3 Field-strength renormalization in φ4 theory

In this problem we evaluate the two-loop corrections to scalar’s two-point function in φ4

theory in the massless limit. There are three diagrams contribute in total.

The first diagram reads

=(−iλ)2

6

∫ddk

(2π)dddq

(2π)di

k2 −m2

i

q2 −m2

i

(p− k − q)2 −m2

=iλ2

6

∫ddkE(2π)d

ddqE(2π)d

i

k2E +m2

i

q2E +m2

i

(pE − kE − qE)2 +m2

=iλ2

6

∫ddkE(2π)d

1

k2E +m2

1

(4π)d/2

∫ 1

0

dxΓ(2− d

2)

[m2 + x(1− x)(pE − kE)2]2−d/2

=iλ2Γ(2− d

2)

6(4π)d/2

∫ 1

0

dx dy

×∫

ddkE(2π)d

[x(1− x)]d/2−2(1− y)1−d/2Γ(3− d2

)/Γ(2− d2

)[(kE − ypE)2 + y(1− y)p2

E +(1− y + y

x(1−x)

)m2]3−d/2

=iλ2

6(4π)d

∫ 1

0

dx dyΓ(3− d)[x(1− x)]d/2−2(1− y)1−d/2[y(1− y)p2

E +(1− y + y

x(1−x)

)m2]3−d . (10.19)

10.4. Asymptotic behavior of diagrams in φ4 theory 83

Now we take m2 = 0 and d = 4− ε→ 4. Then we have

=iλ2

12(4π)4Γ(−1 + ε)(p2

E)1−ε + · · · = − iλ2

12(4π)4p2E

[ 1

ε− log(p2

E) + · · ·]

=iλ2

12(4π)4p2[ 1

ε− log(−p2) + · · ·

]. (10.20)

The second diagram actually vanishes in m→ 0 limit. In fact,

=−iδλ

2

∫ddk

(2π)di

k2 −m2=−iδλ

2(4π)d/2Γ(1− d

2)

m1−d/2 ∝ m→ 0. (10.21)

The third diagram reads ip2δZ . Therefore we can choose the counterterm δZ , under the MS

scheme, to be

δZ = − λ2

12(4π)4

[1

ε− logM2

]. (10.22)

Thus the field strength counterterm receives a nonzero contribution at this order. In the

massless limit, it is

δ2Γ(2) =iλ2

12(4π)4p2 log

M2

−p2. (10.23)

10.4 Asymptotic behavior of diagrams in φ4 theory

In this problem we calculate the four point amplitude in φ4 theory to 2-loop order in

s→∞, t fixed, limit. The tree level result is simply −iλ, and the 1-loop result can be easily

evaluated to be

iδ1M =(−iλ)2

2

∫ddk

(2π)di

k2 −m2

[ i

(ps − k)2 −m2+

i

(pt − k)2 −m2+

i

(pu − k)2 −m2

]− iδλ

' iλ2

2(4π)2

[3(

2ε− γ + log 4π

)− log s− log t− log u

]− iδλ

=− iλ2

2(4π)2

(log s+ log t+ log u

)∼ − iλ2

(4π)2log s (10.24)

In the last step we take the limit s → ∞. In this limit t can be ignored and u ' −s. We

see the divergent part of the counterterm coefficient δλ at 1-loop order is

δλ ∼3λ2

(4π)2

1

ε. (10.25)

Now we consider the two-loop correction.


=(−iλ)3

4

[ ∫ddk

(2π)di

k2

i

(ps − k)2

]2

= − iλ3

4(4π)d

[ ∫ 1

0

dxΓ(2− d

2)

[−x(1− x)s]2−d/2

]2

∼− iλ3

(4π)4

( 1

ε2− 1

εlog s+

1

2log2 s

). (10.26)

In the last line we only keep the divergent terms as ε→ 0 and s→∞.

=(−iλ)3

2

∫ddkddq

(2π)2d

i

k2

i

(ps − k)2

i

q2

i

(k − p3 − q)2

=− iλ3

2

∫ddk

(2π)d1

k2(ps − k)2

[i

(4π)d/2

∫ 1

0

dxΓ(2− d

2)

[x(1− x)(kE − p3E)2]2−d/2

]=

iλ3

2(4π)d/2

∫ 1

0

dxΓ(2− d

2)

[x(1− x)]2−d/2

∫ddkE(2π)d

1

k2E(psE − kE)2[(kE − p3E)2]2−d/2

=iλ3

2(4π)d/2

∫ 1

0

dxΓ(2− d

2)

[x(1− x)]2−d/2

∫ddkE(2π)d

∫ 1

0

dy

∫ 1−y

0

dzz1−d/2

(k2E + ∆)4−d/2

Γ(4− d2

)

Γ(2− d2

)

=iλ3

2(4π)d

∫dxdydz

z1−d/2

[x(1− x)]2−d/2Γ(4− d)

∆4−d , (10.27)

where ∆ = ys+ zp23E − (ypsE + zp3E)2.

Then we find

∼ − iλ3

(4π)4

( 1

ε2− 1

εlog s+

1

2log2 s

). (10.28)

The same result for the third diagram. Then we have

+ + ∼ − iλ3

(4π)4

( 3

ε2− 3

εlog s+

3

2log2 s

). (10.29)

Now we come to the counterterm. The fourth diagram reads

=(−iλ)(−iδλ)

2

∫ddk

(2π)di

k2

i

(ps − k)2

∼ 3λ3

2(4π)2

1

ε

i

(4π)2

2

ε

(1− ε

2log s+

ε2

8log2 s+ · · ·

)∼ 3iλ3

(4π)4

( 1

ε2− 1

2εlog s+

1

8log2 s

)(10.30)

The same result for the fifth diagram. Then we have

+ + + + ∼ iλ3

(4π)4

( 3

ε2− 3

4log2 s

). (10.31)

10.4. Asymptotic behavior of diagrams in φ4 theory 85

So much for the s-channel. The t and u-channel results can be obtained by replacing s with

t and u respectively. In the limit s → ∞ and t-fixed, we can simply ignore t and treating

u ∼ −s, then the total 2-loop correction in this limit is

iδ2M∼ −3iλ3

2(4π)4log2 s. (10.32)

The double pole 1/ε2 has been absorbed by δλ.

In summary, we have the following asymptotic expression for the 4-point amplitude to

2-loop order in the s→∞ and t-fixed limit:

iM = −iλ− iλ2

(4π)2log s− 3iλ3

2(4π)4log2 s+ · · · . (10.33)


Chapter 11

Renormalization and Symmetry

11.1 Spin-wave theory

(a) Firstly we prove the following formula:⟨Teiφ(x)e−iφ(0)

⟩= e[D(x)−D(0)]. (11.1)

Where D(x) = 〈Tφ(x)φ(0)〉 is the time-ordered correlation of two scalars. The left hand

side of this equation can be represented by path integral, as

1

Z[0]

∫Dφ eiφ(x)e−iφ(0) exp

[i

∫ddxddy 1

2φ(x)D−1(x− y)φ(y)

]. (11.2)

This expression precisely has the form Z[J ]/Z[0], with J(y) = δ(y − x) − δ(0). Thus we

have

Z[J ]/Z[0] = − 1

2

∫ddxddy J(x)D(x− y)J(y) = exp

[D(x)−D(0)

], (11.3)

which is just the right hand side of the formula.

(b) The operator being translational invariant O[φ(x)] = O[φ(x) − α] can depend on φ

only through ∇µφ. And the only relevant/marginal Lorentz-invariant operator satisfying

this condition is 12ρ(∇φ)2.

(c) From now on we use bold x to denote coordinate and italic x to denote its length,

x ≡ |x|. We can use the result in (a) to evaluate 〈s(x)s∗(0)〉, as

〈s(x)s∗(0)〉 = A2〈eiφ(x)e−iφ(0)〉 = A2eD(x)−D(0). (11.4)

Note that the correlation function

D(x) =1

ρ

∫ddkE(2π)d

1

k2E

eikE ·x (11.5)

87

88 Chapter 11. Renormalization and Symmetry

is the solution to the following equation:

− ρ∇2D(x− y) = δ(d)(x− y). (11.6)

Since D(x) is a function of the length only, namely D(x) = D(x), thus we have

− ρ

xd−1

∂

∂x

(xd−1 ∂

∂xD(x)

)=

Γ(1 + d2

)

dπd/2δ(x)

xd−1. (11.7)

Then it’s easy to find

D(x) =

Γ(1 + d

2)

d(d− 2)πd/2ρ

1

xd−2, for d 6= 2,

− 1

2πρlog x, for d = 2.

(11.8)

Then we have

Dimension d d = 1 d = 2 d = 3 d = 4

D(x) − 1

2ρx − 1

2πρlog x

1

4πρx

1

4π2ρx2

〈ss∗〉 ∼ e−x ∼ 1/x2πρ ∼ e1/x ∼ e1/x2

Since ρ→ 0 when d→ 2, the correlation function 〈ss∗〉 in this case is independent of length

x.

11.2 A zeroth-order natural relation

We study N = 2 linear sigma model coupled to fermions:

L = 12∂µφ

i∂µφi + 12µ2φiφi − 1

4λ(φiφi)2 + ψ(i/∂)ψ − gψ(φ1 + iγ5φ2)ψ, (11.9)

with φi a two-component field, i = 1, 2.

(a) Now, under the following transformation:

φ1 → φ1 cosα− φ2 sinα; φ2 → φ1 sinα + φ2 cosα; ψ → e−iαγ2/2ψ, (11.10)

the first three terms involving φi only keep invariant. The fourth term, as the kinetic term

of a chiral fermion, is also unaffected by this transformation. Thus, to show the whole

Lagrangian is invariant, we only need to check the last term, and this is really the case:

− gψ(φ1 + iγ5φ2)ψ

→− gψe−iαγ5/2[(φ1 cosα− φ2 sinα) + iγ5(φ1 sinα + φ2 cosα)

]e−iαγ5/2ψ

=− gψe−iαγ5/2eiαγ5(φ1 + iγ5φ2)e−iαγ5/2ψ = −gψ(φ1 + iγ5φ2)ψ. (11.11)

11.2. A zeroth-order natural relation 89

(b) Now let φ acquire a vacuum expectation value v, which equals to√µ2/λ classically.

Then, in terms of new variables φ = (v + σ(x), π(x)), the Lagrangian reads

L = 12

(∂µσ)2 + 12

(∂µπ)2 − µ2σ2 − 14λ(σ4 + π4)

− 12λσ2π2 − λvσ3 − λvσπ2 + ψ(i/∂ − gv)ψ − gψ(σ + iγ5π)ψ. (11.12)

That is, the fermion acquire a mass mf = gv.

(c) Now we calculate the radiative corrections to the mass relation mf = gv. The renor-

malization conditions we need are as follows.

q p

p′π = gγ5 at q2 = 0, p2 = p′2 = m2

f . (11.13)

σ= 0. (11.14)

These two conditions fixed g and v so that they receive no radiative corrections. Then we

want to show that the mass of the fermion mf receives finite radiative correction at 1-loop.

Since the tadpole diagrams of σ sum to zero by the renormalization condition above, the

fermion’s self-energy receive nonzero contributions from the following three diagrams:

The first two 1-loop diagrams can be evaluated as

(e) = (−ig)2

∫ddk

(2π)di

/k −mf

i

(k − p)2 − 2µ2= g2

∫ddk

(2π)d

∫ 1

0

dxx/p+mf

(k′2 −∆1)2

=ig2

(4π)d/2

∫ 1

0

dxΓ(2− d

2)

∆2−d/21

(x/p+mf )

=ig2

(4π)2

∫ 1

0

dx (x/p+mf )[

2ε− γ + log 4π − log ∆1

](11.15)

(f) = g2

∫ddk

(2π)dγ5 i

/k −mf

γ5 i

(k − p)2= g2

∫ddk

(2π)d

∫ 1

0

dxx/p−mf

(k′2 −∆2)2

=ig2

(4π)d/2

∫ 1

0

dxΓ(2− d

2)

∆2−d/22

(x/p−mf )

=ig2

(4π)2

∫ 1

0

dx (x/p−mf )[

2ε− γ + log 4π − log ∆2

]. (11.16)


This leads to

(e) + (f) =ig2

(4π)2

∫ 1

0

dx

2x/p[

2ε− γ + log 4π − 1

2log(∆1∆2)

]+mf log

∆2

∆1

(11.17)

We see that the correction to the fermions mass mf from these two diagrams is finite. Be-

sides, the third diagram, namely the counterterm, contributes the mass’ correction through

δgv. The the total correction to mf is finite only when δg is finite. Let us check this by

means of the first renormalization condition (11.13) stated above. The 1-loop contributions

to (11.13) are as follows.

(a) = (−ig)2g

∫ddk

(2π)di

/k −mf

γ5 i

/k −mf

i

(k − p)2 − 2µ2

= ig3

∫ddk

(2π)d(/k +mf )γ

5(/k +mf )

(k2 −m2f )

2((k − p)2 − 2µ2

) = −ig3γ5

∫ddk

(2π)d1

(k2 −m2f )((k − p)2 − 2µ2

)=− ig3γ5

∫ddk′

(2π)d

∫ 1

0

dx1

(k′2 −∆1)2=

g3γ5

(4π)d/2

∫ 1

0

dxΓ(2− d

2)

∆2−d/21

=g3γ5

(4π)2

∫ 1

0

dx[

2ε− γ + log 4π − log ∆1

](11.18)

(b) = g3

∫ddk

(2π)dγ5 i

/k −mf

γ5 i

/k −mf

γ5 i

(k − p)2 − 2µ2

=− ig3

∫ddk

(2π)dγ5(/k +mf )γ

5(/k +mf )γ5

(k2 −m2f )

2((k − p)2 − 2µ2

) = ig3γ5

∫ddk

(2π)d1

(k2 −m2f )((k − p)2 − 2µ2

)= ig3γ5

∫ddk′

(2π)d

∫ 1

0

dx1

(k′2 −∆2)2=−g3γ5

(4π)d/2

∫ 1

0

dxΓ(2− d

2)

∆2−d/22

=−g3γ5

(4π)2

∫ 1

0

dx[

2ε− γ + log 4π − log ∆2

](11.19)

(c) = (−ig)g(−2iλv)

∫ddk

(2π)dγ5 i

/k −mf

i

(k − p)2 − 2µ2

i

(k − p)2

= 4ig2λvγ5

∫ddk′

(2π)d

∫ 1

0

dx

∫ 1−x

0

dy(x+ y)/p+mf

(k′2 −∆3)3

=2g2λvγ5

(4π)2

∫ 1

0

dx

∫ 1−x

0

dy(x+ y)/p+mf

∆3

(11.20)

11.3. The Gross-Neveu model 91

(d) = (−ig)g(−2iλv)

∫ddk

(2π)di

/k −mf

γ5 i

(k − p)2 − 2µ2

i

(k − p)2

= 4ig2λvγ5

∫ddk′

(2π)d

∫ 1

0

dx

∫ 1−x

0

dy−(x+ y)/p+mf

(k′2 −∆3)3

=2g2λvγ5

(4π)2

∫ 1

0

dx

∫ 1−x

0

dy−(x+ y)/p+mf

∆3

(11.21)

Thus,

(a) + (b) + (c) + (d) =gγ5

(4π)2

∫ 1

0

dx[g2 log

∆2

∆1

+ 4λ

∫ 1−x

0

dym2f

∆3

]. (11.22)

11.3 The Gross-Neveu model

The Gross-Neveu Model is a theory of fermions in 1 + 1 dimensional spacetime:

L = ψii/∂ψi + 12g2(ψiψi)

2, (11.23)

with i = 1, · · · , N . The gamma matrices are taken as γ0 = σ2, γ1 = iσ1, where σi is the

familiar Pauli matrices. We also define γ5 = γ0γ1 = σ3.

(a) The theory is invariant under the transformation ψi → γ5ψi. It is straightforward to

check this. We note that:

ψi = ψ†iγ0 → ψ†iγ

5γ0 = −ψiγ5, (11.24)

thus:

L →− ψiγ5i/∂γ5ψi + 12g2(−ψiγ5γ5ψi)

2

= ψii/∂ψi + 12g2(ψiψi)

2. (11.25)

However, a mass term will transform as miψiψi → −miψiψi, thus a theory respecting this

chiral symmetry does not allow such a mass term.

(b) The superficial renormalizability of the theory (by power counting) is obvious since

[g] = 0.

(c) The model can be phrased in another equivalent way:

Z =

∫DψDψDσ exp

[i

∫d2x

(ψii/∂ψi − 1

2g2σ2 − σψiψi

)]. (11.26)

This can be justified by integrating out σ,∫Dσ exp

[i

∫d2x (− 1

2g2σ2 − σψiψi)

]= N exp

[i

∫d2x g2

2(ψiψi)

2

]. (11.27)

which recovers the following path integral:

Z =

∫DψDψ exp

[i

∫d2x

(ψii/∂ψi + 1

2g2(ψiψi)

2)]. (11.28)


(d) We can also integrate out the fermions ψi to get the effective potential for the auxiliary

field σ: ∫DψDψ exp

[i

∫d2x(ψii/∂ψi − σψiψi

)]=[

det(i/∂ − σ)]N

=[

det(∂2 + σ2)]N

= exp

[ ∫d2k

(2π)2N log(−k2 + σ2)

]. (11.29)

The integral is divergent, which should be regularized. We use the dimensional regulariza-

tion: ∫ddkE(2π)d

N log(k2E + σ2) = N

∫ddkE(2π)d

[∂

∂α

1

k2E + σ2

]α=0

=− iNΓ(−d/2)(σ2)d/2

(4π)d/2. (11.30)

Now we set d = 2− ε and send ε→ 0,∫ddkE(2π)d

N log(k2E + σ2) =

iNσ2

4π

(2

ε− γ + log 4π − log σ2 + 1

). (11.31)

Thus the effective potential is

Veff(σ) =1

2g2σ2 +

N

4πσ2

(log

σ2

µ2− 1

)(11.32)

by modified minimal subtraction.

(e) Now we minimize the effective potential:

0 =∂Veff

∂σ=

1

g2σ +

N

2πσ log

σ2

µ2, (11.33)

and find nonzero vacuum expectation values 〈σ〉 = ±µe−π/g2N . The dependence of this result

on the renormalization condition is totally in the dependence on the subtraction point µ.

(f) It is well-known that the loop expansion is equivalent to the expansion in powers

of ~ in generic perturbation theory around a classical vacuum. This is true because the

integrand of the partition function can be put into the form of eiS/~. That is, ~ appears as

an overall coefficient of the action. In our case, we see that the overall factor N plays the

same role. Thus by the same argument, we conclude that the loop expansion is equivalent

to the expansion in powers of 1/N . More details can be found in Section III.3 of [4] and

Chapter 8 “1/N” of [5].

Chapter 12

The Renormalization Group

12.1 Beta Function in Yukawa Theory

In this problem we calculate the 1-loop beta functions in Yukawa theory. All needed

ingredients have been given in Problem 10.2 Here we list the needed counterterms:

δψ = − g2

2(4π2)

(2

ε− logM2

); (12.1)

δφ = − 2g2

(4π2)

(2

ε− logM2

); (12.2)

δg =g3

(4π)2

(2

ε− logM2

); (12.3)

δλ =3λ2 − 48g4

2(4π)2

(2

ε− logM2

). (12.4)

Here Λ is the UV cutoff and M is the renormalization scale. Then, the beta functions to

lowest order are given by

βg = M∂

∂M

(− δg + 1

2g0δφ + g0δψ

)=

5g3

(4π)2; (12.5)

βλ = M∂

∂M

(− δλ + 2λ0δφ

)=

3λ2 + 8λg2 − 48g4

(4π)2. (12.6)

12.2 Beta Function of the Gross-Neveu Model

We evaluate the β function of the 2-dimensional Gross-Neveu model with the Lagrangian

L = ψi(i/∂)ψi + 12g2(ψiψi)

2, (i = 1, · · · , N) (12.7)

to 1-loop order. The Feynman rules can be easily worked out to be

ki α j β =

( i

/k

)βαδij

93

94 Chapter 12. The Renormalization Group

i α j β

k γ l δ

= ig2(δijδk`εβαεδγ + δi`δjkεδαεβγ)

Now consider the two-point function Γ(2)ij (p). The one-loop correction to Γ

(2)ij (p) comes from

the following two diagrams:

i j + i j

It is easy to see the loop diagram contains a factor of∫

d2k tr [/k−1

], which is zero un-

der dimensional regularization. Thus the wave function renormalization factor receives no

contribution at 1-loop level, namely δψ = 0.

Then we turn to the 4-point function Γ(4)ijk`. There are three diagrams in total, namely,

γ′ δ′

α′ β′

(a)

nm

iα jβ

kγ `δ

+

(b)

n

miα

kγ

jβ

`δ

+

(c)

m

n

iα

`δ

jβ

kγ

We calculate them in turn. The first one:

(a) = (ig2)2

∫ddk

(2π)d(δmnδk`εδγεγ′δ′ + δn`δmkεδδ′εγ′γ)

( i

/k

)δ′β′

× (δijδmnεβ′α′εβα + δinδjmεβ′αεβα′)( i

/k

)α′γ′

= g4(

(−2N + 2)δijδk`εδγεβα + 12δi`δjn(γµ)δα(γµ)βγ

)∫ ddk

(2π)d1

k2(12.8)

The second diagram reads:

(b) =1

2· (ig2)2

∫ddk

(2π)d(δmjδn`εδδ′εββ′ + δm`δnjεδβ′εβδ′)

( i

/k

)β′α′

× (δimδknεα′αεγ′γ + δinδkmεγ′αεα′γ)( i

−/k

)δ′γ′

=− g4

2

(δijδk`(γ

µ)δγ(γµ)βα + δi`δjn(γµ)δα(γµ)βγ

)∫ ddk

(2π)d1

k2(12.9)

The third diagram:

(c) = (ig2)2

∫ddk

(2π)d(δmjδnkεββ′εγ′γ + δmnδjkεγ′β′εβγ)

( i

/k

)β′α′

× (δimδ`nεδδ′εα′α + δi`δmnεα′δ′εδα)( i

/k

)δ′γ′

12.3. Asymptotic Symmetry 95

=g4(

12δijδk`(γ

µ)δγ(γµ)βα + (2− 2N)δi`δjkεβγεδα

)∫ ddk

(2π)d1

k2(12.10)

Summing up the three diagrams and using dimensional regularization with d = 2 − ε, we

get

− 2g4(N − 1)(δijδk`εδγεβα + δi`δjkεβγεδα)

∫ddk

(2π)d1

k2

∼ 2(N − 1)ig4

4π

2

ε(δijδk`εδγεβα + δi`δjkεβγεδα). (12.11)

Only the divergent terms are kept in the last expression, from which we can read the

counterterm

δg = − (N − 1)g4

2π

( 2

ε− logM2

). (12.12)

Thus the β function is

β(g2) = M∂

∂M(−δg) = − (N − 1)(g2)2

π, (12.13)

and

β(g) = − (N − 1)g3

2π. (12.14)

It is interesting to see that the 1-loop β function vanishes for N = 1. This is because we

have the Fierz identity 2(ψψ)(ψψ) = −(ψγµψ)(ψγµψ), and the Gross-Neveu model in this

case is equivalent to massless Thirring model, which is known to have vanishing β function.

12.3 Asymptotic Symmetry

In this problem we study a bi-scalar model, given by the following Lagrangian:

L = 12

((∂µφ1)2 + (∂µφ2)2

)− λ

4!(φ4

1 + φ42)− ρ

12φ2

1φ22. (12.15)

(a) First, we calculate the 1-loop beta functions βλ and βρ. The relevant 1-loop diagrams

for calculating βλ are:

The relevant diagrams for calculating βρ are:


Here the single line represents φ1 and double line represents φ2. Since the divergent parts

of these diagrams are all independent of external momenta, we can therefore simply ignore

them. Then it’s easy to evaluate them, as follows.

=(−iλ)2

2

∫d4k

(2π)4

i

k2

i

k2∼ iλ2

2(4π)2

2

ε, (12.16)

=(−iρ/3)2

2

∫d4k

(2π)4

i

k2

i

k2∼ iρ2

18(4π)2

2

ε. (12.17)

The t-channel and u-channel give the same result. Thus we can determine δλ to be

δλ ∼9λ2 + ρ2

6(4π)2

2

ε. (12.18)

On the other hand,

= =(−iλ)(−iρ/3)

2

∫d4k

(2π)4

i

k2

i

k2∼ iλρ

6(4π)2

2

ε, (12.19)

= = (−iρ/3)2

∫d4k

(2π)4

i

k2

i

k2∼ iρ2

9(4π)2

2

ε. (12.20)

Then we have

δρ ∼3λρ+ 2ρ2

3(4π2)

2

ε. (12.21)

It’s easy to see that field strengths for both φ1 and φ2 receives no contributions from 1-loop

diagrams. Thus the 1-loop beta functions can be evaluated as

βλ = −µ dδλdµ

=3λ2 + ρ2/3

(4π)2; (12.22)

βρ = −µ dδρdµ

=2λρ+ 4ρ2/3

(4π)2. (12.23)

(b) Now we derive the renormalization equation for ρ/λ:

µd

dµ

( ρλ

)=

1

λβρ −

ρ

λ2βλ =

ρ

3(4π)2

[− (ρ/λ)2 + 4(ρ/λ)− 3

]. (12.24)

Then it is easy to see that ρ/λ = 1 is an IR fixed point.

(c) In 4− ε dimensions, the β functions for ρ and λ are shifted as

βλ = −ελ+3λ2 + ρ2/3

(4π)2; (12.25)

βρ = −ερ+2λρ+ 4ρ2/3

(4π)2. (12.26)

But it is easy to show that the terms containing ε cancel out in the β function for ρ/λ, and

the result is the same as (12.24). This is true because ρ/λ still remains dimensionless in

4− ε dimensions. Therefore we conclude that there are three fixed points of the RG flow for

ρ/λ at 0, 1, and 3. We illustrate this in the diagram of RG flow in the ρ-λ plane, with the

deviation of dimension ε = 0.01, in Figure 12.1.

12.3. Asymptotic Symmetry 97

0.0 0.2 0.4 0.6 0.8 1.0

0.0

0.2

0.4

0.6

0.8

1.0

Λ

Ρ

Figure 12.1: The RG flow of the theory (12.15) in 4 − ε dimensions with ε = 0.01. Three

nontrivial fixed points are shown by blue dots.


Chapter 13

Critical Exponents and Scalar Field

Theory

13.1 Correlation-to-scaling exponent

In this problem we consider the effect of the deviation of the coupling λ from its fixed

point to the two-point correlation function G(M, t) in d = 4 − ε dimensions. Symbolically,

we can always write

G(M, t) = G∗(M, t) +δG(M, t)

δλ

∣∣∣∣λ=λ∗

δλ, (13.1)

where λ is the running coupling, defined to be the solution of the following renormalization

group equation:d

d log µλ =

2βλ(λ)

d− 2 + 2γ(λ). (13.2)

As the first step, let us expand the β function of λ around the fixed point, as

β(λ) = β(λ∗) +dβ(λ)

dλ

∣∣∣∣λ=λ∗

(λ− λ∗) +O((λ− λ∗)2)

= ω(λ− λ∗) +O((λ− λ∗)2). (13.3)

Then the renormalization group equation reads

d

d log µλ ' 2ω(λ− λ∗)

d− 2 + 2γ(λ∗)=

ων

β(λ− λ∗), (13.4)

where β and ν on the right hand side are critical exponents, which in our case are defined

to be

β =d− 2 + 2γ(λ∗)

d(2− γφ∗(λ∗)), ν =

1

2− γφ2(λ∗).

Don’t confuse the critical exponent β with the β function. Now, from this equation we can

solve the running coupling λ to be

λ = λ∗ +(λ(µ0)− λ∗

)( µµ0

)ων/β. (13.5)

99

100 Chapter 13. Critical Exponents and Scalar Field Theory

Now let µ0 be the scale at which the bare coupling is defined. Then we get

δλ ∝ (λ− λ∗)µων/β. (13.6)

13.2 The exponent η

We have found the counterterm δZ to O(λ2) with MS scheme in Problem 10.3, to be

δZ = − λ2

12(4π)4

[1

ε− logM2

]. (13.7)

Then the anomalous dimension γ to O(λ2) is given by

γ =1

2M

∂

∂MδZ =

λ2

12(4π)4. (13.8)

This result can be easily generalized to the O(N)-symmetric φ4 theory, by replacing the

Feynman rule of the φ4 coupling −iλ with

−2iλ(δijδk` + δikδj` + δi`δjk),

which is equivalent to multiplying the diagram (10.19) by the following factor:

4 ·(δikδ`m + δi`δkm + δimδk`

)(δjkδ`m + δj`δkm + δjmδk`

)= 12(N + 2)δij, (13.9)

and the anomalous dimension (13.8) obtained above should be multiplied by 12(N + 2),

which leads to

γ = (N + 2)λ2

(4π)4, (13.10)

which is the same as (13.47) of Peskin&Schroeder.

13.3 The CPN model

(a) The Lagrangian of the CPN model can be written as

L =1

g2

(∑j

|∂µzj|2 −∣∣∣∑

j

z∗j∂µzj

∣∣∣2), (13.11)

with zj (j = 1, · · ·N + 1) the components of a vector in (N + 1) dimensional complex space,

subject to the constraint ∑j

|zj|2 = 1 (13.12)

and the identification

(eiαz1, · · · , eiαzN+1) ∼ (z1, · · · , zN+1). (13.13)

13.3. The CPN model 101

Now we prove that the Lagrangian given above is invariant under the following local trans-

formation:

zj(x)→ eiα(x)zj(x), (13.14)

as,

g2L →∣∣∂µ(eiαzj)

∣∣2 +∣∣e−iαz∗j∂µ(eiαzj)

∣∣2=(|∂µzj|2 + |∂µα|2 + 2Re

(− i(∂µα)z∗j∂µzj

))−(|z∗j∂µzj|2 + |∂µα|2 + 2Re

(− i(∂µα)ziz

∗i z∗j∂µzj

))= g2L. (13.15)

Then we show that the nonlinear σ model with n = 3 is equivalent to the CPN model

with N = 1. To see this, we substitute ni = z∗σiz into the Lagrangian of the nonlinear

sigma model, L = 12g2|∂µni|2, to get

L =1

2g2

∣∣∣(∂µz∗)σiz + z∗σi∂µz∣∣∣2

=1

2g2σiσi

[2(∂µz

∗)(∂µz)z∗z + (∂µz)2z∗2 + (∂µz∗)2z2

]=

1

2g2σiσi

[2(∂µz

∗)(∂µz) +(z∗∂µz + z∂µz

∗)2 − 2(z∗∂µz)(z∂µz∗)]

=1

2g2σiσi

[2(∂µz

∗)(∂µz) +[∂µ(z∗z)]2 − 2(z∗∂µz)(z∂µz∗)

]. (13.16)

Then after a proper normalization of the field z, it is straightforward to see that the La-

grangian above reduces to

L =1

g2

(|∂µz|2 − 2|z∗∂µz|2

), (13.17)

which is indeed the CP 1 model.

(b) The Lagrangian (13.11) can be obtained by the following Lagrangian with a gauge field

Aµ and a Lagrange multiplier which expresses the local gauge symmetry and the constraint

explicitly:

L =1

g2

(|Dµzj|2 − λ

(|zj|2 − 1

)), (13.18)

with Dµ = ∂µ + iAµ. Now let us verify this by functionally integrating out the gauge field

Aµ as well as the Lagrange multiplier λ to get

Z =

∫D2ziDAµDλ exp

[i

g2

∫d2x

(|Dµzj|2 − λ

(|zj|2 − 1

))]=

∫D2ziDAµ δ

(|zj|2 − 1

)exp

[i

g2

∫d2x |Dµzj|2

]=

∫D2ziDAµ δ

(|zj|2 − 1

)exp

[i

g2

∫d2x

(AµA

µ + 2iAµ(∂µz∗j )zj + |∂µzj|2

)]=N

∫D2zi δ

(|zj|2 − 1

)exp

[i

g2

∫d2x

(|∂µzj|2 − |z∗j∂µzj|

)2]. (13.19)

102 Chapter 13. Critical Exponents and Scalar Field Theory

(c) On the other hand one can also integrate out zi field in the Lagrangian (13.18), as

Z =

∫DziDAµDλ exp

[i

g2

∫d2x

(|Dµzj|2 − λ

(|zj|2 − 1

))]=

∫DAµDλ exp

[−N tr log(−D2 − λ) +

i

g2

∫d2xλ

](13.20)

We assume that the expectation values for Aµ and λ are constants. Then the exponent can

be evaluated by means of dimensional regularization, as

iS = −N tr log(−D2 − λ) +i

g2

∫d2xλ

=

[−N

∫ddk

(2π)dlog(k2 + AµA

µ − λ)

+i

g2λ

]· V (2)

⇒ i

[− N

4π

(log

M2

λ− A2+ 1)

(λ− A2) +1

g2λ

]· V (2), (13.21)

where V (2) =∫

d2x, and we have used the MS scheme to subtract the divergence. Now we

can minimize the quantity in the square bracket in the last line to get

Aµ = 0, λ = M2 exp(− 4π

gN2

). (13.22)

(d) The meaning of the effective action S is most easily seen from its diagrammatic rep-

resentations. For instance, at the 1-loop level, we know that the logarithmic terms in the

effective action is simply the sum of a series of 1-loop diagrams with n ≥ 0 external legs,

where the number of external legs n is simply the power of corresponding fields in the expan-

sion of S. Therefore, to the second order in A and in λ, the effective action is represented

precisely by the following set of diagrams,

where the dashed lines represent λ, curved lines represent Aµ, and the internal loop are

z field. Then it is straightforward to see that the correct kinetic terms for λ and Aµ are

generated from these diagrams. That is, the gauge field Aµ becomes dynamical due to

quantum corrections. The gauge invariance of the resulted kinetic term FµνFµν can also be

justified by explicit calculation as was done in Problem 9.1.

Final Project II

The Coleman-Weinberg Potential

In this final project, we work out some properties of Coleman-Weinberg model, illustrat-

ing basic techniques of the renormalization group. The original paper [6] by S. Coleman

and E. Weinberg is always a good read, while a recent and very insightful treatment of the

model can be found in [7].

Simply put, the Coleman-Weinberg model is a theory of scalar electrodynamics, described

by the Lagrangian,

L = − 12FµνF

µν + (Dµφ)†(Dµφ)−m2φ†φ− λ6

(φ†φ)2, (13.23)

with φ a complex scalar and Dµφ = (∂µ + ieAµ)φ.

(a) Consider the case of spontaneous breaking of the U(1) gauge symmetry φ(x) →eiα(x)φ(x), caused by a negative squared mass, namely m2 = −µ2 < 0. The scalar then

acquires a nonzero vacuum expectation value (VEV) φ0 =√〈|φ|2〉. We split this VEV out

of the scalar field, namely,

φ = φ0 +1√2

[σ(x) + iπ(x)

], (13.24)

with the new field σ(x) and π(x) being real. At the tree level, it is easy to find φ0 =√

3µ2/λ

by minimize the scalar potential V (φ) = −µ2φ†φ + λ6

(φ†φ)2. We also introduce v =√

2φ0

for convenience. Then, rewrite the Lagrangian in terms of these new field variables, we get,

L =− 14

(Fµν)2 + 1

2(∂µσ)2 + 1

2(∂µπ)2 + 1

2e2v2AµA

µ − 12

(2µ2)σ2

− λ24

(π4 + σ4 + 2π2σ2 + 4vπ2σ + 4vσ3) + evAµ∂µπ

+ eAµ(σ∂µπ − π∂µσ) + 12e2AµA

µ(π2 + σ2 + 2vσ). (13.25)

Then we see that the vector field Aµ acquires a mass, equal to mA = ev at the classical

level.

(b) Now we calculate the 1-loop effective potential of the model. We know that 1-loop

correction of the effective Lagrangian is given by,

∆L =i

2log det

[− δ2Lδϕδϕ

]ϕ=0

+ δL, (13.26)

103

104 Final Project II. The Coleman-Weinberg Potential

where ϕ is the fluctuating fields and δL denotes counterterms.

Let the background value of the complex scalar be φcl. By the assumption of Poincare

symmetry, φcl must be a constant. For the same reason, the background value of the vector

field Aµ must vanish. In addition, we can set φcl to be real without loss of generality. Then

we have,

φ(x) = φcl + ϕ1(x) + iϕ2(x),

where ϕ1(x), ϕ2(x), together with Aµ(x), now serve as fluctuating fields. Expanding the

Lagrangian around the background fields and keeping terms quadratic in fluctuating fields

only, we get,

L =− 12FµνF

µν +∣∣(∂µ + ieAµ)(φcl + ϕ1 + iϕ2)

∣∣2−m2

∣∣φcl + ϕ1 + iϕ2

∣∣2 − λ6

∣∣φcl + ϕ1 + iϕ2

∣∣4= 1

2Aµ[gµν(∂2 + 2e2φ2

cl)− ∂µ∂ν]Aν + 1

2ϕ1

(− ∂2 −m2 − λφ2

cl

)ϕ1

+ 12ϕ2

(− ∂2 −m2 − λ

3φ2

cl

)ϕ2 − 2eφclAµ∂

µϕ2 + · · · , (13.27)

where “· · · ” denotes terms other than being quadratic in fluctuating fields. Now we impose

the Landau gauge condition ∂µAµ = 0 to the Lagrangian, which removes the off-diagonal

term −2eφclAµ∂µϕ2. Then, according to (13.26), the 1-loop effective Lagrangian can be

evaluated as,

i

2log det

[− δ2Lδϕδϕ

]ϕ=0

=i

2

[log det

(− ηµν(∂2 + 2e2φ2

cl) + ∂µ∂ν)

+ log det(∂2 +m2 + λφ2

cl

)+ log det

(∂2 +m2 + λ

3φ2

cl

)]=

i

2

∫ddk

(2π)d

[tr log(−k2 + 2e2φ2

cl)3

+ tr log(−k2 +m2 + λφ2cl) + tr log(−k2 +m2 + λ

3φ2

cl)

]=

Γ(− d2

)

2(4π)d/2

[3(2e2φ2

cl)d/2 + (m2 + λφ2

cl)d/2 + (m2 + λ

3φ2

cl)d/2

]. (13.28)

In the second equality we use the following identity,

det(λI + AB) = λn−1(λ+BA), (13.29)

where A and B are matrices of n × 1 and 1 × n, respectively, λ is an arbitrary complex

number and I is the n× n identity matrix. In our case, this gives,

det(− ηµν(∂2 + 2e2φ2

cl) + ∂µ∂ν)

= −2e2φ2cl(∂

2 + 2e2φ2cl)

3. (13.30)

Then the second equality follows up to an irrelevant constant term. The third equality

makes use of the trick in (11.72) of P&S. Then, for d = 4− ε and ε→ 0, we have,

i

2log det

[− δ2Lδϕδϕ

]ϕ=0

=1

4(4π)2

[3(2e2φ2

cl)2(∆− log(2e2φ2

cl))


+ (m2 + λφ2cl)

2(∆− log(m2 + λφ2

cl))

+ (m2 + λ3φ2

cl)2(∆− log(m2 + λ

3φ2

cl))], (13.31)

where we define ∆ ≡ 2ε− γ + log 4π + 3

2for brevity.

Now, with MS scheme, we can determine the counterterms in (13.26) to be

δL =−1

4(4π)2

[2

ε− γ + log 4π − logM2

](3(2e2φ2

cl)2 + (m2 + λφ2

cl)2 + (m2 + λ

3φ2

cl)2).

(13.32)

where M is the renormalization scale. Now the effective potential follows directly from

(13.26), (13.31) and (13.32),

Veff[φcl] = m2φ2cl +

λ

6φ4

cl −1

4(4π)2

[3(2e2φ2

cl)2(

logM2

2e2φ2cl

+3

2

)+ (m2 + λφ2

cl)2(

logM2

m2 + λφ2cl

+3

2

)+ (m2 + λ

3φ2

cl)2(

logM2

m2 + λ3φ2

cl

+3

2

)]. (13.33)

(c) Now taking the mass parameter µ2 = −m2 = 0, then the effective potential (13.33)

becomes

Veff[φcl] =λ

6φ4

cl +1

4(4π)2

[12e4φ4

cl

(log

2e2φ2cl

M2− 3

2

)+

10

9λ2φ4

cl

(log

λφ2cl

M2− 3

2

)]' λ

6φ4

cl +3e4φ4

cl

(4π)2

(log

2e2φ2cl

M2− 3

2

). (13.34)

In the second line we use the fact that λ is of the order e4 to drop the λ2 term. Then the

minimal point of this effective potential can be easily worked out to be,

φ2cl =

M2

2e2exp

(1− 8π2λ

9e4

). (13.35)

As λ ∼ e4, we see that φcl is of the same order with e−1M . Thus the effective potential

remains valid at this level of perturbation theory.

(d) We plot the effective potential as a function of φcl in Figure (13.1). The purple curve

with m2 = 5× 10−7M2 corresponds the case with no spontaneous symmetry breaking. The

blue curve shows that as m2 goes to 0 from above, new local minima is formed. Finally, the

orange and red curves correspond to broken symmetry, and in the case of the orange curve

with m2 = 0, the symmetry is dynamically broken.

(e) Now we calculate β functions of the Coleman-Weinberg model to 1-loop level at high

energies, where we can send the mass parameter m2 to zero. It is convenient to work in the

Feynman gauge ξ = 1. Then the relevant Feynman rules can be read from the Lagrangian

(13.25) to be,


0 1 2 3 4 5 6 7

-0.06

-0.04

-0.02

0.00

0.02

ΦclH10-2

M L

Veff@Φ

clDH

10-

2ML

Figure 13.1: The effective potential Veff as a function of φcl, with different values of m2/M2 =

5× 10−7, 2.4× 10−7, 0 and −1× 10−7 from top to bottom, respectively.

=i

k2, =

i

k2, =

−iηµνk2

,

= −iλ, = −iλ, = − iλ

3,

= 2ie2ηµν , = 2ie2ηµν = e(k1 − k2)µ

We first find the 1-loop wave function renormalization. For σ field, there is only one

diagram with nonzero contribution,

p−→

p− k−→

which reads,

e2

∫ddk

(2π)di

k2

−i

(p− k)2(p+ k)µ(−p− k)µ ∼ − 2ie2p2

(4π)2

2

ε. (13.36)

Then we have,

δσ =2e2

(4π)2

( 2

ε−M2

), (13.37)


and it is straightforward to see that δπ = δσ. For photon’s wave function renormalization

(vacuum polarization), we need to evaluate the following three diagrams,

The sum of the three diagrams is,

− e2

∫ddk

(2π)di

k2

i

(p− k)2(2k − p)µ(2k − p)ν + 2 · 1

2· 2ie2

∫ddk

(2π)di

k2· (p− k)2

(p− k)2

∼ − ie2

3(4π)2

2

ε(p2ηµν − pµpν), (13.38)

which gives,

δA = − e2

3(4π)2

( 2

ε− logM2

). (13.39)

Then we turn to the 1-loop corrections to couplings. For scalar self-coupling λ, we

consider the 1-loop corrections to σ4 term in the Lagrangian. There are six types of diagrams

contributing, listed as follows, and we label them by (a) to (f) from left to right,

For each type there are several different permutations of internal lines giving identical result,

or more concretely, 3 permutations for each of the first three types, and 6 permutations for

each of the last three types. Now we evaluate them in turn. We set all external momenta

to zero to simplify the calculation. Then,

(a) =(−iλ)2

2

∫ddk

(2π)d

( i

k2

)2

∼ iλ2

2(4π)2

2

ε, (13.40)

(b) =(−iλ/3)2

2

∫ddk

(2π)d

( i

k2

)2

∼ iλ2

18(4π)2

2

ε, (13.41)

(c) =(2ie2)2

2

∫ddk

(2π)d

( −i

k2

)2

ηµνηµν ∼ 8ie4

(4π)2

2

ε, (13.42)

(d) =−iλe2

3

∫ddk

(2π)d−i

k2

( i

k2

)2

(−kµkµ) ∼ − iλe2

3(4π)2

2

ε, (13.43)

(e) = (2ie2)e2

∫ddk

(2π)di

k2

( −i

k2

)2

(−kµkµ) ∼ − 2ie4

(4π)2

2

ε, (13.44)

(f) = e4

∫ddk

(2π)d

( i

k2

)2( −i

k2

)2

(−kµkµ)2 ∼ ie4

(4π)2

2

ε, (13.45)


Then multiplying (a)∼(c) by 3 and (d)∼(f) by 6, we find

δλ =5λ2/3− 2λe2 + 18e4

(4π)2

2

ε. (13.46)

Finally we consider the 1-loop corrections to e. For this purpose we calculate 1-loop diagrams

with three external lines with 1 Aµ, 1 σ and 1 π respectively, shown as follows, labeled again

by (a) to (d) from left to right,

Now we calculate them in turn.

(a) = e3

∫ddk

(2π)d−i

k2

( i

(p− k)2

)2

(2p− k)2(2k − 2p)µ

= ie3

∫ddk′

(2π)d

∫ 1

0

dx4x[(2− x)p− k′]2[k′ − (1− x)p]µ

[k′2 + x(1− x)p2]3

∼ ie3

∫ddk′

(2π)d

∫ 1

0

dx−k′2(1− x)pµ − 2(2− x)(p · k′)k′µ

[k′2 + x(1− x)p2]3

= ie3

∫ddk′

(2π)d

∫ 1

0

dx[−(1− x)− 2

d(2− x)]k′2pµ

[k′2 + x(1− x)p2]3

∼ 2e3

(4π)2

2

εpµ, (13.47)

(b) = e( −iλ

3

)∫ ddk

(2π)d

( i

k2

)2

· 2kµ = 0, (13.48)

(c) = (d) = e(2ie2)

∫ddk

(2π)di

k2

−i

(p− k)2(k + p)µ ∼ − 3e3

(4π)2

2

εpµ. (13.49)

Summing the four diagrams, we find that

δe =2e3

(4π)2

( 2

ε− logM2

). (13.50)

Now we are ready to calculate β functions,

βe = M∂

∂M

(− δe +

1

2(δA + δσ + δπ)

)=

e3

48π2, (13.51)

βλ = M∂

∂M

(− δλ + 2δσ

)=

5λ2 − 18λe2 + 54e4

24π2. (13.52)

The trajectory of renormalization group flows generated from these β functions are shown

in Figure 13.2.


0.0 0.1 0.2 0.3 0.4 0.5

0.0

0.1

0.2

0.3

0.4

0.5

Λ

e2

Figure 13.2: The renormalization group flow of Coleman-Weinberg model.

(f) The effective potential obtained in (c) is not a solution to the renormalization group

equation, since it is only a first order result in perturbation expansion. However, it is possible

to find an effective potential as a solution to the RG equation, with the result in (c) serving

as a sort of “initial condition”. The effective potential obtained in this way is said to be RG

improved.

The Callan-Symansik equation for the effective potential reads(M

∂

∂M+ βλ

∂

∂λ+ βe

∂

∂e− γφφcl

∂

∂φcl

)Veff(φcl, λ, e;M) = 0. (13.53)

The solution to this equation is well known, that is, the dependence of the sliding energy

scale M is described totally by running parameters,

Veff(φcl, λ, e;M) = Veff

(φcl(M

′), λ(M ′), e(M ′);M ′), (13.54)

where barred quantities satisfy

M∂λ

∂M= βλ(λ, e), M

∂e

∂M= βe(λ, e), M

∂φcl

∂M= −γφ(λ, e)φcl. (13.55)

The RG-improved effective potential should be such that when expanded in terms of coupling

constants λ and e, it will recover the result in (c) at the given order. For simplicity here

we work under the assumption that λ ∼ e4, so that all terms of higher orders of coupling

constants than λ and e4 can be ignored. In this case, the perturbative calculation in (c)

gives

Veff =λ

6φ4

cl +3e4φ4

cl

(4π)2

(log

2e2φ2cl

M2− 3

2

). (13.56)


Now we claim that the RG-improved edition of this result reads

Veff =λ

6φ4

cl +3e4φ4

cl

(4π)2

(log 2e2 − 3

2

). (13.57)

To see this, we firstly solve the renormalization group equations (13.55),

λ(M ′) = e4

(λ

e4+

9

4π2log

M ′

M

), (13.58)

e2(M ′) =e2

1− (e2/24π2) log(M ′/M), (13.59)

φcl(M′) = φcl

(M ′

M

)2e2/(4π)2

, (13.60)

where the unbarred quantities λ, e and φcl are evaluated at scale M . Now we substitute

these results back into the RG-improved effective potential (13.57) and expand in terms of

coupling constants. Then it is straightforward to see that the result recovers (13.56). To see

the spontaneous symmetry breaking still occurs, we note that the running coupling λ(M ′)

flows to negative value rapidly for small M ′ = φcl, while e(M ′) changes mildly along the

φcl scale, as can be seen directly from Figure 13.2. Therefore the the coefficient before φ4cl

is negative for small φcl and positive for large φcl. As a consequence, the minimum of this

effective potential should be away from φcl = 0, namely the U(1) symmetry is spontaneously

broken.

To find the scalar mass mσ in this case (with µ = 0), we calculate the second derivative

of the effective potential Veff with respect to φcl. Since the renormalization scale M can

be arbitrarily chosen, we set it to be M2 = 2e2〈φ2cl〉 to simplify the calculation. Then the

vanishing of the first derivative of Veff at φcl = 〈φcl〉 implies that λ = 9e4/8π2. Insert this

back to Veff in (13.56), we find that

Veff =3e4φ4

cl

16π2

(log

φ2cl

〈φ2cl〉− 1

2

). (13.61)

Then, taking the second derivative of this expression with respect to φcl, we get the scalar

mass m2σ = 3e4〈φ2

cl〉/4π2 = 3e4v2/8π2. Recall that the gauge boson’s mass mA is given by

mA = e2v2 at the leading order, thus we conclude that m2σ/m

2A = 3e2/8π2 at the leading

order in e2.

(g) Now we consider the effect of finite mass, by adding a positive quadratic term into the

effective potential (13.56). For simplicity we still work with λ = 9e4/8π2, which is always

attainable without tuning. Then the effective potential reads,

Veff = m2rφ

2cl +

3e4φ4cl

(4π)2

(log

2e2φ2cl

M2− 1

2

), (13.62)

in which the mass mr is not identical to the bare mass parameter appeared in the classical

Lagrangian, but has include 1-loop correction. As mr increases from zero, the energy of


symmetry breaking vacuum at 〈φcl〉 6= 0 also increases, until it reaches zero when m =

mc > 0, and has the same vacuum energy with the symmetric vacuum 〈φcl〉 = 0. Then for

m > mc, there will be no stable symmetry breaking vacuum.

Using the effective potential above, we can find the position of the symmetry breaking

vacuum by solving the equation ∂Veff/∂φcl

∣∣φcl=φv

= 0, with the following solution,

φ2v =

M2

2e2exp

[W(− (4π)2m2

r

3e2M2

)], (13.63)

in which W(z) is the Lambert W function, defined as the solution of z = W(z)eW(z). Then

we can use the condition Veff(φv) = 0 to determine mc to be,

m2c =

3e2M2

32π2e1/2, (13.64)

where the e in denominator is 2.718... and should not be confused with electric charge e.

Now we can evaluate the mass ratio m2σ/m

2A = 1

2V ′′eff(φv)/(eφv)

2, as a function of mass

parameter mr, to be,

m2σ/m

2A =

3e2

8π2

[1 + W

(− (4π)2m2

r

3e2M2

)]. (13.65)

This is a monotonically decreasing function of mr, and when mr = 0, it recovers the previous

result 3e2/8π2. On the other hand, when mr = mc, the mass ratio m2σ/m

2A reaches its

minimum value, given by 3e2

8π2 [1 + W(− 12e

)] = 3e2/16π2, which is one half of the massless

case.

(h) When the spacetime dimension is shifted from 4 as d = 4− ε, the β functions βe and

βλ are also shifted to be

βe = −εe+e3

48π2, βλ = −ελ+

5λ2 − 18λe2 + 54e4

24π2. (13.66)

We plot the corresponding RG flow diagrams for several choice of ε in Figure 13.3, where

we also extrapolate the result to ε = 1.


0.0 0.1 0.2 0.3 0.4 0.5

0.0

0.1

0.2

0.3

0.4

0.5

Λ

e2

0.0 0.1 0.2 0.3 0.4 0.5

0.0

0.1

0.2

0.3

0.4

0.5

Λ

e2

0.0 0.1 0.2 0.3 0.4 0.5

0.0

0.1

0.2

0.3

0.4

0.5

Λ

e2

0.0 0.1 0.2 0.3 0.4 0.5

0.0

0.1

0.2

0.3

0.4

0.5

Λ

e2

Figure 13.3: The renormalization group flows of Coleman-Weinberg model in d = 4 − ε

spacetime dimensions, with ε = 0.005, 0.01, 0.1 and 1 in the upper-left, upper-right, lower-

left and lower-right diagram, respectively.

Chapter 15

Non-Abelian Gauge Invariance

15.1 Brute-force computations in SU(3)

(a) The dimension of SU(N) group is d = N2 − 1, when N = 3 we get d = 8.

(b) It’s easy to see that t1, t2, t3 generate a SU(2) subgroup of SU(3). Thus we have

f ijk = εijk for i, j, k = 1, 2, 3. Just take another example, let’s check [t6, t7]:

[t6, t7] = i(− 12t3 +

√3

2t8),

thus we get

f 678 =√

32, f 673 = − 1

2.

Then what about f 376?

[t3, t7] = i2t6 ⇒ f 376 = 1

2= −f 673.

(c) C(F ) = 12

. Here F represents fundamental representation.

(d) C2(F ) = 43

, d(F ) = 3, d(G) = 8, thus we see that d(F )C2(F ) = d(G)C(F ).

15.2 Adjoint representation of SU(2)

The structure constants for SU(2) is fabc = εabc, thus we can write down the represen-

tation matrices for its generators directly from

(tbG)ac = ifabc = iεabc.

More explicitly,

t1G =

0 0 0

0 0 −i

0 i 0

, t2G =

0 0 i

0 0 0

−i 0 0

, t3G =

0 −i 0

i 0 0

0 0 0

, (15.1)

113

114 Chapter 15. Non-Abelian Gauge Invariance

Then,

C(G) = tr (t1Gt1G) = tr (t2Gt

2G) = tr (t3Gt

3G) = 2,

C2(G)I3 = t1Gt1G + t2Gt

2G + t3Gt

3G = 2I3 ⇒ C2(G) = 2.

Here I3 is the 3× 3 unit matrix.

15.3 Coulomb potential

(a) We calculate vacuum expectation value for Wilson loop UP (z, z), defined by

UP (z, z) = exp

[− ie

∮P

dxµAµ(x)

]. (15.2)

By definition, we have

〈UP (z, z)〉 =

∫DAµ exp

[iS[Aµ]− ie

∮P

dxµAµ(x)

], (15.3)

where

S[Aµ] =

∫d4x

[− 1

4FµνF

µν − 12ξ

(∂µAµ)2]. (15.4)

Thus 〈UP (z, z)〉 is simply a Gaussian integral, and can be worked out directly, as

〈UP (z, z)〉 = exp

[− 1

2

(− ie

∮P

dxµ)(− ie

∮P

dyν) ∫ d4k

(2π)4

−igµνk2 + iε

e−ik·(x−y)

](15.5)

Here we have set ξ → 0 to simplify the calculation. Working out the momentum integral,

we get

〈UP (z, z)〉 = exp

[− e2

8π2

∮P

dxµ∮P

dyνgµν

(x− y)2

]. (15.6)

The momentum integration goes as follows∫d4k

(2π)4

e−ik·(x−y)

k2 + iε= i

∫d4kE(2π)4

eikE ·(x−y)

−k2E

=i

(2π)4

∫ 2π

0

dψ

∫ π

0

dφ sinφ

∫ π

0

dθ sin2 θ

∫ ∞0

dkE k3E

eikE |x−y| cos θ

−k2E

= − i

4π3

∫ ∞0

dkE kE

∫ π

0

dθ sin2 θeikE |x−y| cos θ

= − i

4π2

∫ ∞0

dkE kEJ1

(kE|x− y|

)kE|x− y|

= − i

4π2(x− y)2. (15.7)

Where J1(x) is Bessel function and we use the fact that∫∞

0dx J1(x) = 1.

15.4. Scalar propagator in a gauge theory 115

(b) Now taking a narrow rectangular Wilson loop P with width R in x1 direction (0 <

x1 < 1) and length T in x0 direction (0 < x0 < T ) and evaluate 〈UP 〉. When the integral

over dx and dy go independently over the loop, divergence will occur as |x− y|2 → 0. But

what we want to show is the dependence of 〈UP 〉 on the geometry of the loop, namely the

width R and length T , which should be divergence free. Therefore, when T R, the

integral in Wilson loop is mainly contributed by time direction and can be expressed as

〈UP (z, z)〉 ' exp

[− 2e2

8π2

∫ T

0

dx0

∫ 0

T

dy0 1

(x0 − y0)2 −R2 − iε

], (15.8)

and we have add a small imaginary part to the denominator for the reason that will be clear.

Carry out the integration, we find∫ T

0

dx0

∫ 0

T

dy0 1

(x0 − y0)2 −R2 − iε

TR−−−→ 2T

Rarctanh

( T

R + iε

)= − i

π

T

R.

Therefore,

〈UP 〉 = exp( ie2

4πR· T)

= e−iV (R)T , (15.9)

which gives the familiar result V (R) = −e2/4πR.

(c) For the Wilson loop of a non-Abelian gauge group, we have

UP (z, z) = tr

P exp

[− ig

∮P

dxµAaµ(x)tar

], (15.10)

where tar is the matrices of the group generators in representation r. We expand this expres-

sion to the order of g2,

UP (z, z) = tr (1)− g2

∮P

dxµ∮P

dyνAaµ(x)Abν(y) tr (tartbr) +O(g3)

= tr (1)

[1− g2C2(r)

∮P

dxµ∮P

dyνAaµ(x)Abν(y)

]+O(g3). (15.11)

Compared with the Abelian case, we see that to order g2, the non-Abelian result is given by

making the replacement e2 → g2C2(r). Therefore we conclude that V (R) = −g2C2(r)/4πR

in non-Abelian case.

15.4 Scalar propagator in a gauge theory

In this problem we study very briefly the heat kernel representation of Green func-

tions/propagator of a scalar field living within a gauge field background.


(a) To begin with, we consider the simplest case, in which the background gauge field

vanishes. Then we can represent the Green function DF (x, y) of the Klein-Gordon equation,

defined to be

(∂2 +m2)DF (x, y) = −iδ(4)(x− y) (15.12)

with proper boundary conditions, by the following integral over the heat kernel function

D(x, y, T ):

DF (x, y) =

∫ ∞0

dT D(x, y, T ). (15.13)

The heat kernel satisfies the following “Schrodinger equation”:[i∂

∂T− (∂2 +m2)

]D(x, y, T ) = iδ(T )δ(4)(x− y). (15.14)

The solution to this equation can be represented by

D(x, y, T ) = 〈x|e−iHT |y〉 =

∫d4k

(2π)4

d4k′

(2π)4〈x|k〉〈k|e−iHT |k′〉〈k′|y〉

=

∫d4k

(2π)4

d4k′

(2π)4e−i(−k2+m2)T e−ik·x+ik′·y(2π)4δ(4)(k − k′)

=

∫d4k

(2π)4ei(k2−m2)T e−ik·(x−y), (15.15)

with H = ∂2 +m2. Integrating this result over T , with the +iε prescription, we recover the

Feynman propagator for a scalar field:∫ ∞0

dT D(x, y, T ) =

∫d4k

(2π)4e−ik·(x−y)

∫ ∞0

dT ei(k2−m2+iε)T

=

∫d4k

(2π)4

ie−ik·(x−y)

k2 −m2 + iε. (15.16)

(b) Now let us turn on a background Abelian gauge field Aµ(x). The corresponding

“Schrodinger equation” then becomes[i∂

∂T−((∂µ − ieAµ(x)

)2+m2

)]D(x, y, T ) = iδ(T )δ(4)(x− y), (15.17)

the solution of which, 〈x|e−iHT |y〉, can also be expressed as a path integral,

〈x|e−iHT |y〉 = limN→∞

∫ N∏i=1

(dxi⟨xi∣∣ exp

− i∆t

[(∂µ − ieAµ(x)

)2+m2

]∣∣xi−1

⟩), (15.18)

where we have identify x = xN , y = x0, and ∆t = T/N . Then,

〈xi|e−i∆t[(∂µ−ieAµ(x))2+m2]|xi−1〉

=

∫d4ki(2π)4

〈xi|e−i∆t[∂2−ieAµ(x)∂µ+m2]|ki〉〈ki|e−i∆t[−ie∂µAµ(x)−e2A2(x)]|xi−1〉

15.5. Casimir operator computations 117

=

∫d4ki(2π)4

〈xi|e−i∆t[−k2i+eAµ(xi)kµi +m2]|ki〉〈ki|e−i∆t[ekµi Aµ(xi−1)−e2A2(xi−1)]|xi−1〉

=

∫d4ki(2π)4

e−i∆t[−k2i+eki·(A(xi)+A(xi−1))−e2A2(xi−1)+m2−iε]e−iki·(xi−xi−1)

= C exp

[− i∆t

4

( xi − xi−1

∆t+ eA(xi) + eA(xi−1)

)2

− i∆t(m2 − e2A2(xi−1))

]⇒ C exp

[− i∆t

4

( dx

dt

)2

− i∆teA(x) · dx

dt− i∆tm2

]. (15.19)

In the last line we take the continuum limit, and C is an irrelevant normalization constant.

Then we get

D(x, y, T ) =

∫Dx exp

[− i

∫ T

0

dt

(( dx

dt

)2

+m2

)− ie

∫ T

0

dx(t) · A(x(t))

]. (15.20)

15.5 Casimir operator computations

(a) In the language of angular momentum theory, we can take common eigenfunctions of

J2 =∑

a TaT a and Jz = T 3 to be the representation basis. Then the representation matrix

for T 3 is diagonal:

t3j = diag (−j,−j + 1, · · · , j − 1, j).

Thus

tr (t3j t3j) =

j∑m=−j

m2 = 13j(j + 1)(2j + 1).

Then we have

tr (t3rt3r) =

∑i

tr (t3jit3ji

) = 13

∑i

ji(ji + 1)(2ji + 1) = C(r),

which implies that

3C(r) =∑i

ji(ji + 1)(2ji + 1). (15.21)

(b) Let the SU(2) subgroup be spanned by T 1, T 1 and T 3. Then in fundamental repre-

sentation, the representation matrices for SU(2) subgroup of SU(N) can be taken as

tiN =

(τi/2 02×(N−2)

0(N−2)×2 0(N−2)×(N−2)

). (15.22)

Where τi (i = 1, 2, 3) are Pauli matrices. We see that the representation matrices for SU(2)

decomposes into a doublet and (N − 2) singlet. Then it’s easy to find that

C(N) = 13

(12

( 12

+ 1)(2 · 12

+ 1))

= 12, (15.23)

by formula in (a).


In adjoint representation, the representation matrices (ti)ab = ifaib (a, b = 1, · · · , N2 −1, i = 1, 2, 3). Thus we need to know some information about structure constants. Here

we give a handwaving illustration by analyzing the structure of fundamental representation

matrices a little bit more. Note that there’re three types of representation matrices, listed

as follows. For convenience, let’s call them tA, tB and tC :

tA =

(A2×2 02×(N−2)

0(N−2)×2 0(N−2)×(N−2)

). (15.24)

tB =

(A2×2 B2×(N−2)

B†(N−2)×2 0(N−2)×(N−2)

). (15.25)

tC =

(− 1

2tr (C)I2×2 02×(N−2)

0(N−2)×2 C(N−2)×(N−2)

). (15.26)

In which, tA is just the representation matrices for SU(2) subgroup. Thus we see that there

are 3 tA, 2(N−2) tB and (N−2)2 tC in total. It’s also obvious that there is no way to generate

a tA from commutators between two tC or between a tB and tC , the only way to generate

tA are commutators between two tA or between to tB. Then, tA commutators correspond to

the triplet representation os SU(2) subgroup, and 2(N − 2)-tB commutators correspond to

the doublet representation of SU(2). In this way we see that adjoint representation matrices

for SU(2) subgroup decompose into 1 triplet, 2(N − 1) doublets and (N − 2)2 singlets.

Then we can calculate C(G), again, by using formula in (a), as:

C(G) = 13

[1(1 + 1)(2 · 1 + 1) + 2(N − 2) · 1

2( 1

2+ 1)(2 · 1

2+ 1)

]= N. (15.27)

(c) Let U ∈ SU(N) be N × N unitary matrix, S be a symmetric N × N matrix, and A

be an antisymmetric N ×N matrix. Then we can use S and A to build two representations

for SU(N) respectively, as

S → USUT , A→ UAUT .

It’s easy to verify that they are indeed representations. Let’s denote these two representation

by s and a. It’s also obvious to see that the dimensions of s and a are d(s) = N(N + 1)/2

and d(a) = N(N − 1)/2 respectively.

Accordingly, the generator T a acts on S and A as:

S → T aS + S(T a)T , A→ T aA+ A(T a)T . (15.28)

To get C2(s) and C2(a), we can make use of the formula

d(r)C2(r) = d(G)C(r). (15.29)

Thus we need to calculate C(r) and C(a). By formula in (a), we can take an generator in

SU(2) subgroup to simplify the calculation. Let’s take

t3N = 12

diag(1,−1, 0, · · · , 0),

15.5. Casimir operator computations 119

Then we have:

S =

S11 · · · S1n

.... . .

...

Sn1 · · · Snn

→ t3NS + S(t3N)T =1

2

2S11 0 S13 · · · S1n

0 2S22 S23 · · · S2n

S31 S32 0 · · · 0...

......

. . ....

Sn1 Sn2 0 · · · 0

A =

0 A12 · · · A1n

A21 0. . .

......

. . . . . . An−1,n

An1 · · · An,n−1 0

→ t3NA+A(t3N)T =1

2

0 0 A13 · · · A1n

0 0 A23 · · · A2n

A31 A32 0 · · · 0...

......

. . ....

An1 An2 0 · · · 0

Thus we see that the representation matrices for T 3, in both s representation and a repre-

sentation, are diagonal. They are:

t3s = diag(1, 0, 12· · · , 1

2︸︷︷︸N−2

, 1, 12, · · · , 1

2︸︷︷︸N−2

, 0, · · · , 0︸︷︷︸(N−2)(N−1)/2

); (15.30)

t3a = diag(0, 12, · · · , 1

2︸︷︷︸2(N−2)

, 0, · · · , 0︸︷︷︸(N−2)(N−3)/2

). (15.31)

Here we have rearrange the upper triangular elements of S and A by line.

Then we get

C(s) = tr (t3s)2 = 1

2(N + 2); (15.32)

C(a) = tr (t3a)2 = 1

2(N − 2). (15.33)

Then,

C2(s) =d(G)C(s)

d(s)=

(N2 − 1)(N + 2)/2

N(N + 1)/2=

(N − 1)(N + 2)

N; (15.34)

C2(a) =d(G)C(a)

d(a)=

(N2 − 1)(N − 2)/2

N(N − 1)/2=

(N + 1)(N − 2)

N. (15.35)

At last let’s check the formula implied by (15.100) and (15.101):(C2(r1) + C2(r2)

)d(r1)d(r2) =

∑C2(ri)d(ri), (15.36)

in which the tensor product representation r1×r2 decomposes into a direct sum of irreducible

representations ri. In our case, the direct sum of representation s and a is equivalent to the

tensor product representation of two copies of N . That is,

N ×N ∼= s+ a.


Thus, we have,

(C2(N) + C2(N)

)d(N)d(N) =

[ N2 − 1

2N+N2 − 1

2N

]N2 = N(N2 − 1);

and

C2(s)d(s) + C2(a)d(a) =[C(s) + C(a)

]d(G) = N(N2 − 1).

Thus formula (15.36) indeed holds in our case.

Chapter 16

Quantization of Non-Abelian Gauge

Theories

16.1 Arnowitt-Fickler gauge

In this problem we perform the Faddeev-Popov quantization of Yang-Mills theory in

Arnowitt-Fickler gauge (also called axial gauge), namely A3a = 0. More generally, we may

write the gauge condition as nµAµa = 0 with nµ an arbitrary space-like vector of unit norm

(n2 = −1). The condition A3a = 0 corresponds simply to the choice nµ = gµ3. This gauge

has the advantage that the Faddeev-Popov ghosts do not propagate and do not couple to

gauge fields, as we will show below.

Our starting point, the partition function, reads

Z =

∫DAµ δ(n · Aa)eiS[Aµ] det

( δα(n · Aa)∂αb

), (16.1)

with S = −14

∫d4x (F a

µν)2 the classical action for the gauge field, and the Faddeev-Popov

determinant is given by

det( δα(n · Aa)

∂αb

)= det

( 1

gnµ∂

µδab − fabcnµAµc)

=

∫DbDc exp

[i

∫d4x ba

(nµ∂

µδab − fabcnµAµc)cb]. (16.2)

When multiplied by the delta function δ(n · Aa), the second term in the exponent above

vanishes, which implies that the ghost and antighost do not interact with gauge field. Mean-

while, they do not propagator either, since there does not exist a canonical kinetic term for

them. Therefore we can safely treat the Faddeev-Popov determinant as an overall normal-

ization of the partition function and ignore it. Then, the partition function reduces to

Z = limξ→0

∫DAµ exp

[i

∫d4x

(− 1

4(F a

µν)2 − 1

2ξ(n · Aa)2

)]= lim

ξ→0

∫DAµ exp

[i

∫d4x

(12Aaµ(gµν∂2 − ∂µ∂ν − 1

ξnµnν

)Abν

121

122 Chapter 16. Quantization of Non-Abelian Gauge Theories

− gfabc(∂κAaλ)AκbAλc − 14g2f eabf ecdAaκA

bλA

κcAλd)]. (16.3)

where we have convert the delta function δ(n · Aa) into a limit of Gaussian function. Then

we see that the three-point or four-point gauge boson vertices share the same Feynman rules

with the ones in covariant gauge. The only difference arises from the propagator. Let us

parameterize the propagator in momentum space as

Dµν(k) = Ak2gµν +Bkµkν + C(kµnν + nνkµ) +Dnµnν . (16.4)

Then, the equation of motion satisfied by the propagator,(gµνk2 − kµkν − 1

ξnµnν

)Dνλ(k) = gµλ (16.5)

gives

A = − i

k2, B = − 1 + ξk2

k · nC, C = − 1

k · nA, D = 0. (16.6)

Note that the gauge fixing parameter ξ should be sent to 0. Therefore the propagator reads

D(k)µν = − i

k2

(gµν +

kµkν

(k · n)2− kµnν + kνnµ

k · n

). (16.7)

16.2 Scalar field with non-Abelian charge

(a) Firstly we write down the Lagrangian for the Yang-Mills theory with charged scalar

field, as

L = − 1

4(F a

µν)2 + (Dµφ)†(Dµφ), (16.8)

where the covariant derivative Dµφ =(∂µ + igAaµt

ar

)φ with tar the matrices of gauge group

generators in representation r. For simplicity we ignore the possible mass term for the scalar.

Then, it is straightforward to derive the Feynman rules for this theory by expanding this La-

grangian. The rules for the propagator and self-interactions of gauge boson are independent

of matter content and are the same with the ones given in Figure 16.1 in Peskin&Schroeder.

The only new ingredients here are the gauge boson-scalar field interactions, which generate

the following Feynman rules:µ, a ν, b

= ig2(tartbr + tbrt

ar)g

µν ,

p1 p2

µ, a

= −igtar(p1 − p2)µ,

16.3. Counterterm relations 123

(b) To compute the β function of coupling g, we introduce some additional Feynman rules

involving counterterms:

p1 p2

aµ

= −igδ1tar(p1 − p2)µ,

p → = ip2δ2 − iδm,

p →aµ bν

= −iδabδ3(p2gµν − pµpν).

Then the β function is given by

β = gM∂

∂M

(− δ1 + δ2 +

1

2δ3

). (16.9)

To determine the counterterms, we evaluate the following relevant 1-loop diagrams. But the

calculations can be simplified a lot if we observe that the combination δ1− δ2 is determined

by pure gauge sector, and is independent of matter content. This may be most easily seen

from the counterterm relation δ1− δ2 = δc1− δc2, where the right hand side comes from ghost

contribution which is a pure gauge quantity. We will demonstrate this counterterm relation

explicitly in the next problem for fermionic matter. Therefore, we can borrow directly the

result of Peskin & Schroeder, or from the result of Problem 16.3(a),

δ1 − δ2 = − g2

(4π)2C2(G)

( 2

ε− logM2

). (16.10)

On the other hand, δ3 can be found by evaluating the loop corrections to the gauge boson’s

self-energy. The contributions from the gauge boson loop and ghost loop have already been

given in eq.(16.71) in Peskin&Schroeder, while the rest of the contributions is from the

scalar-loop, and is simply the result we have found in Problem 9.1(c), multiplied by the

gauge factor tr (tartbr) = C(r) and the number of scalar ns. Combining these two parts gives

the divergent part of δ3:

δ3 =g2

(4π)2

[ 5

3C2(G)− 1

3nsC(r)

]( 2

ε− logM2

). (16.11)

Then it is straightforward to see that

β = − g3

(4π)2

( 11

3C2(G)− 1

3nsC(r)

). (16.12)

16.3 Counterterm relations

In this problem we calculate the divergent parts of counterterms in Yang-Mills theory

with Dirac spinors at 1-loop level, to verify the counterterm relations, which is a set of


constraints set by gauge invariance. To begin with, let us rewrite the Lagrangian in its

renormalized form, with counterterms separated, as

L =− 14

(∂µAaν − ∂νAaµ)2 + ψ(i/∂ −m)ψ − ca∂2ca

+ gAaµψγµtaψ − gfabc(∂µAaν)AbµAcν

− 14g2(f eabAaµA

bν)(f

ecdAcµAdν)− gcafabc∂µ(Abµcc)

− 14δ3(∂µA

aν − ∂νAaµ)2 + ψ(iδ2/∂ − δm)ψ − δc2ca∂2ca

+ gδ1Aaµψγ

µψ − gδ3g1 f

abc(∂µAaν)AbµA

cν

− 14δ4g

1 (f eabAaµAbν)(f

ecdAcµAdν)− gδc1cafabc∂µ(Abµcc). (16.13)

Then the counterterm relations we will verify are

δ1 − δ2 = δ3g1 − δ3 = 1

2(δ4g

1 − δ3) = δc1 − δc2. (16.14)

Note that δ1 and δ2 have been given in (16.84) and (16.77) in Peskin&Schroeder. Here we

simply quote the results:

δ1 =− g2

(4π)2

[C2(r) + C2(G)

]( 2

ε− logM2

), (16.15)

δ2 =− g2

(4π)2C2(r)

( 2

ε− logM2

). (16.16)

Therefore,

δ1 − δ2 = − g2

(4π)2C2(G)

( 2

ε− logM2

). (16.17)

(a) Firstly let us check the equality δ1− δ2 = δc1− δc2. The 1-loop contributions to δc1 come

from the following three diagrams:

p1k

p2

c a

bµ

The first diagram reads

(−g)3fadef ebff fdc∫

ddk

(2π)d−i

k2

i

(p1 − k)2

i

(p2 − k)2· pν2(p2 − k)µ(p1 − k)ν

⇒− g3fadef ebff fdc∫

ddk

(2π)dkµ(p2 · k)

k6= − 1

4g3fadef ebff fdcpµ2

∫ddk

(2π)d1

k4

⇒− i

4g3fadef ebff fdcpµ2 ·

i

(4π)2

2

ε. (16.18)


The second diagram reads

(−g)2gfadef ebff fdc∫

ddk

(2π)di

k2

−i

(p1 − k)2

−i

(p2 − k)2

× p2ρkσ[gµρ(k − p2 − q)σ + gµσ(q − p1 + k)ρ + gσρ(p1 + p2 − 2k)µ

]⇒− ig3fadef ebff fdc

∫ddk

(2π)d1

k6

[pµ2k

2 + kµ(k · p2)− 2kµ(k · p2)]

⇒− 3

4ig3fadef ebff fdcpµ2

∫ddk

(2π)d1

k4⇒ − 3

4ig3fadef ebff fdcpµ2 ·

i

(4π)2

2

ε. (16.19)

To simplify the structure constant product, we make use of the Jacobi identity,

0 = f ebf (fabdfdce + f bcdfdae + f cadfdbe) = 2fabdfdcef ebf − f cafC2(G),

then we have

fadef ebff fdc = −1

2fabcC2(G). (16.20)

Note that the third diagram reads −gδc1fabcpµ2 , thus we see that to make the sum of these

three diagrams finite, the counterterm coefficient δc1 should be

δc1 ∼ −g2C2(G)

2(4π)2

( 2

ε− logM2

). (16.21)

Then consider δc2. This coefficient should absorb the divergence from the following dia-

gram:

p ka b

This diagram reads

(−g)2f bcdfdca∫

ddk

(2π)di

k2

−i

(p− k)2(p · k)

=− g2C2(G)δab∫

ddk′

(2π)d

∫ 1

0

dxp · (k′ + xp)

(k′2 −∆)2

⇒− g2

2C2(G)δabp2 · i

(4π)2

2

ε+ terms indep. of p2. (16.22)

The corresponding counterterm contributes iδc2p2, therefore we have

δc2 ∼g2C2(G)

2(4π)2

( 2

ε− logM2

). (16.23)

Combining (16.21), (16.23) and (16.17), we see that the equality δ1−δ2 = δc1−δc2 is satisfied.


(b) Now let’s verify the equality δ1 − δ2 = δ3g1 − δ3. In this case the calculation turns

out to be more cumbersome, though. The coefficient δ3 has been given by (16.74) in Pe-

skin&Schroeder. The result is

δ3 =g2

(4π)2

[ 5

3C2(G)− 4

3nfC(r)

]( 2

ε− logM2

). (16.24)

Thus we only need to calculate δ3g1 . The relevant loop diagrams are listed as follows.

p

p ↓

bν cρ

aµ

For simplicity, we have set the external momenta to be p, −p and 0 for the three external

gauge boson lines labeled with (aµ), (bν) and (cρ). Then the contribution of the counterterm

to this vertex is given by

gδ3g1 f

abc(2gµνpρ − gνρpµ − gρµpν). (16.25)

To extract the divergent part from δ3g1 , we have to evaluate the loop diagrams shown above.

Let us calculate them now in turn. The first diagram reads

= 12g(−ig2)fade

[fdeff bcf (gλνgκρ − gλρgκν)

+ fdbff ecf (gλκgνρ − gλρgκν) + fdcff ebf (gλκgνρ − gλνgκρ)]

×∫

ddk

(2π)d−i

k2

−i

(k − p)2

[gµλ(p+ k)κ + gλκ(−2k + p)µ + gµκ(k − 2p)λ

]= 1

2ig3fabcC2(G) · 3

2(gλνgκρ − gλρgκν)

×∫

ddk

(2π)d1

k2

1

(k − p)2

[gµλ(p+ k)κ + gλκ(−2k + p)µ + gµκ(k − 2p)λ

]⇒ 1

2ig3fabcC2(G) · 9

2(gµνpρ − gµρpν)

∫ddk

(2π)d1

k4

⇒ 9

4ig3fabcC2(G)(gµνpρ − gµρpν) · i

(4π)2

2

ε. (16.26)

There are two additional diagrams associated with this diagram by the two cyclic permuta-

tions of the three external momenta. One gives

9

4ig3fabcC2(G)(gµνpρ − gνρpµ) · i

(4π)2

2

ε,

while the other yields zero. Therefore the sum of these three diagrams gives:

9

4ig3fabcC2(G)(2gµνpρ − gµρpν − gνρpµ) · i

(4π)2

2

ε, (16.27)


Then we come to the second diagram, which reads

= g3fadff bedf cfe∫

ddk

(2π)d−i

k2

−i

k2

−i

(p+ k)2

×[gµσ(2p+ k)κ + gσκ(−p− 2k)µ + gκµ(k − p)σ

]×[gνλ(−p+ k)σ + gλσ(−2k − p)ν + gνσ(2p+ k)λ

]×[gρκkλ − 2gκλk

ρ + gρλkκ]

⇒ ig3[− 1

2fabcC2(G)

] ∫ ddk′

(2π)d

∫ 1

0

dx2(1− x)k′2

(k′2 −∆)3

× 1

4

[2(8 + 15x)gµνpρ + (30x− 23)

(gµρpν + gνρpµ

)]⇒ ig3

[− 1

2fabcC2(G)

]· 13

4

(2gµνpρ − gµρpν − gνρpµ

) ∫ ddk′

(2π)d1

k′4

⇒− 13

8ig3fabcC2(G)

(2gµνpρ − gµρpν − gνρpµ

)· i

(4π)2

2

ε. (16.28)

The third diagram reads

= (−g)3fdaff ebdf fce∫

ddk

(2π)d(−1)

( i

k2

)2 i

(k + p)2· (p+ k)µkνkρ

=− ig3 · 12fabcC2(G)

∫ddk′

(2π)d

∫ 1

0

dx2(1− x)k′2

(k′2 −∆)3

× 1d

[− xgµνpρ − xgµρpν + (1− x)gνρpµ

]⇒ 1

24ig3fabcC2(G)(gµνpρ + gµρpν − 2gνρpµ) · i

(4π)2

2

ε. (16.29)

There is again a similar diagram with ghost loop running reversely, which gives

1

24ig3fabcC2(G)(gµνpρ − 2gµρpν + gνρpµ) · i

(4π)2

2

ε.

Then these two diagrams with ghost loops sum to

1

24ig3fabcC2(G)(2gµνpρ − gµρpν − gνρpµ) · i

(4π)2

2

ε. (16.30)

Finally we consider the fourth diagram with fermion loop. There are also two copies with

fermions running in opposite directions. One (shown in the figure) gives

= nf (ig)3 tr (tatctb)

∫ddk

(2π)d(−1) tr

[γµ

i

/kγρ

i

/kγν

i

/k + /p

]⇒ 4

3nfg

3 tr (tatctb)(2gµνpρ − gνρpµ − gµρpν) · i

(4π)2

2

ε, (16.31)

while the other gives

− 4

3nfg

3 tr (tatbtc)(2gµνpρ − gνρpµ − gµρpν) · i

(4π)2

2

ε.


Thus they sum to

4

3nfg

3 tr(ta[tc, tb]

)(2gµνpρ − gνρpµ − gµρpν) · i

(4π)2

2

ε

=− 4

3infg

3C(r)fabc(2gµνpρ − gνρpµ − gµρpν) · i

(4π)2

2

ε. (16.32)

Now, sum up the four groups of diagrams, we get

g3

(4π)2

2

εfabc

[(− 9

4+

13

8− 1

24

)C2(G) +

4

3nfC(r)

](2gµνpρ − gνρpµ − gµρpν), (16.33)

and consequently,

δ3g1 =

g2

(4π)2

[ 2

3C2(G)− 4

3nfC(r)

]·( 2

ε− logM2

). (16.34)

Thus,

δ3g1 − δ3 = − g2

(4π)2C2(G)

( 2

ε− logM2

), (16.35)

which equals to δ1 − δ2 (16.17), as expected.

(c) Now let’s move to the relation δ1 − δ2 = 12

(δ4g1 − δ3). This time we have to evaluate

δ4g1 , which is determined by the divergent part of the following five types of diagrams:

cρ dσ

aµ bν

Firstly the counterterm itself contributes to the 1-loop corrections with

− δ4g1

[fabef cde(gµρgνσ − gµσgνρ) + facef bde(gµνgρσ − gµσgνρ)

+ fadef bce(gµνgρσ − gµρgνσ)]. (16.36)

To evaluate the loop diagrams, we set all external momenta to zero for simplicity. The first

diagram then reads

=1

2(−ig2)2

[fabgf efg(gµλgνκ − gµκgνλ) + faegf bfg(gµνgλκ − gµκgνλ)

+ fafgf beg(gµνgλκ − gµλgνκ)][f efhf cdh(gρλg

σκ − gλσgκρ )

+ f echf fdh(gλκgρσgλσg

ρκ) + f edhf fch(gλκg

ρσ − gρλgσκ)] ∫ ddk

(2π)d

( −i

k2

)2

⇒ ig4

2(4π)2

2

ε

[fabgf efgf efhf cdh(2gµρgνσ − 2gµσgνρ)


+ fabgf efgf echf fdh(gµρgνσ − gµσgνρ) + fabgf efgf edhf fch(gµσgνρ − gµρgνσ)

+ faegf bfgf efhf cdh(gµρgνσ − gµσgνρ) + fafgf begf efhf cdh(gµσgνρ − gµρgνσ)

+ tr (tatbtdtc)(2gµνgρσ + gµρgνσ) + tr (tatbtctd)(2gµνgρσ + gµσgνρ)

+ tr (tatbtctd)(2gµνgρσ + gµσgνρ) + tr (tatbtdtc)(2gµνgρσ + gµρgνσ)]

=ig4

2(4π)2

2

ε

[tr (tatbtctd)(4gµνgρσ − 8gµρgνσ + 10gµσgνρ)

+ tr (tatbtdtc)(4gµνgρσ + 10gµρgνσ − 8gµσgνρ)], (16.37)

where we have used (16.20) and fabc = i(ta)bc with ta the generators in adjoint representation.

There are two additional diagrams similar to this one, which can be obtained by exchange

of labels as (bν ↔ cρ) and (bν ↔ dσ). Therefore the total contribution from these three

diagrams is

ig4

(4π)2

2

ε


+ tr (tatbtdtc)(7gµνgρσ + 7gµρgνσ − 8gµσgνρ)

+ tr (tatctbtd)(−8gµνgρσ + 7gµρgνσ + 7gµσgνρ)]. (16.38)

The second diagram has five additional counterparts. The one displayed in the figure

reads

= (−ig2)g2faegf bgf[f efhf cdh(gρλg

σκ − gσλgρκ)

+ f echf fdh(gλκgρσ − gσλgκρ ) + f edhf fch(gλκg

ρσ − gλρgσκ)]

×∫

ddk

(2π)d

( −i

k2

)3

(gµλkτ − 2kµ + gτµkλ)(gντ kκ − 2gκτ k

ν + gκνkτ )

= g4faegf bgf[f efhf cdh(gρλg

σκ − gσλgρκ)

+ f echf fdh(gλκgρσ − gσλgκρ ) + f edhf fch(gλκg

ρσ − gλρgσκ)]

×∫

ddk

(2π)d

( 1

k2

)2(2gµλg

νκ + 5gµνgλκ − 4gµκg

νλ

)⇒ g4

4

i

(4π)2

2

ε

[faegf bgff efhf cdh(6gµρgνσ − gµσgνρ)

+ faegf bgff echf fdh(13gµνgρσ + 4gµρgνσ − 2gµσgνρ)

+ faegf bgff edhf fch(13gµνgρσ − 2gµρgνσ + 4gµσgνρ)]

=g4

4

i

(4π)2

2

ε

[if cdh tr (tatbth)(6gµρgνσ − gµσgνρ)

− tr (tatbtdtc)(13gµνgρσ + 4gµρgνσ − 2gµσgνρ)

− tr (tatbtctd)(13gµνgρσ − 2gµρgνσ + 4gµσgνρ)]

=− g4

4

i

(4π)2

2

ε



+ tr (tatbtdtc)(13gµνgρσ + 10gµρgνσ − 8gµσgνρ)], (16.39)

where we have used the fact that faegf bgff efhf cdh = if cdh tr (tatbth) = tr(tatb[tc, td]

), and

faegf bgff echf fdh = − tr (tatbtdtc), etc. There are additional five diagrams associated with

this one, namely,

cρ dσ

aµ bν(a)

cρ bν

aµ dσ(b)

aµ dσ

cρ bν(c)

dσ bν

cρ aµ(d)

aµ cρ

bν dσ(e)

In the diagrams above, (a) gives the identical result to the one have just evaluated, while (b)

and (c) give identical expression, so do (d) and (e). We can find (b) from the result above

by the exchange (aµ ↔ cρ), and (d) by the exchange (aµ ↔ dσ). Then we sum up all six

diagrams, which is equivalent to summing the original one with (b) and (d) and multiplying

the result by 2:

− 2 · ig4

4(4π)2

2

ε



+ tr (tctbtatd)(13gµσgνρ − 8gµρgνσ + 10gµνgρσ)

+ tr (tctbtdta)(13gµσgνρ − 8gµνgρσ + 10gµρgνσ)

+ tr (tdtbtcta)(13gµρgνσ − 8gµνgρσ + 10gµσgνρ)

+ tr (tdtbtatc)(13gµρgνσ − 8gµσgνρ + 10gµνgρσ)]

=− ig4

2(4π)2

2

ε



+ tr (tatctbtd)(−16gµνgρσ + 23gµρgνσ + 23gµσgνρ)], (16.40)

where we use the cyclic symmetry of trace and also the relation tr (tatbtctd) = tr (tdtbtcta).

The third diagram reads

= g4faehf bhgf cfefdgf∫

ddk

(2π)d

( −i

k2

)4

(gµλkη − 2gληkµ + gηµkλ)

× (gνηkξ − 2gξηk

ν + gξνkη)(gρκkλ − 2gκλk

ρ + gρλkκ)

× (gσξ kκ − 2gξκkσ + gσκkξ)

= g4 tr (tatbtdtc)

∫ddk

(2π)d1

(k2)4

[34kµkνkρkσ + k4(gµνgρσ + gµρgνσ)

+ 3k2(gµνkρkσ + gµρkνkσ + gνσkµkρ + gρσkµkν)]

= g4 tr (tatbtdtc)

∫ddk

(2π)d1

(k2)2

[ 34

24(gµνgρσ + gµρgνσ + gµσgνρ)


+ (gµνgρσ + gµρgνσ) +3

2(gµνgρσ + gµρgνσ)

]=

ig4

12(4π)2

2

εtr (tatbtdtc)(47gµνgρσ + 47gµρgνσ + 17gµσgνρ). (16.41)

Combined with the other two similar diagrams, we get

ig4

12(4π)2

2

ε

[tr (tatbtctd)(47gµνgρσ + 17gµρgνσ + 47gµσgνρ)

+ tr (tatbtdtc)(47gµνgρσ + 47gµρgνσ + 17gµσgνρ)

+ tr (tatctbtd)(17gµνgρσ + 47gµρgνσ + 47gµσgνρ)]

(16.42)

The fourth diagram with ghost loop is given by

= (−g)4f eahfhbgf gdff fce∫

ddk

(2π)d(−1)

( i

k2

)4

kµkνkρkσ

⇒− ig4

24(4π)2

2

εtr (tatbtdtc)(gµνgρσ + gµρgνσ + gµσgνρ). (16.43)

There are six distinct diagrams with ghost loops, with different permutations of external

labels (Lorentz and gauge). They sum to

− ig4

12(4π)2

2

ε

[tr (tatbtctd) + tr (tatbtdtc) + tr (tatctbtd)

]× (gµνgρσ + gµρgνσ + gµσgνρ). (16.44)

Finally the diagram with fermion loop reads

= (ig)4nf tr (tartbrtdrtcr)

∫ddk

(2π)d(−) tr

[γµ

i

/kγµ

i

/kγσ

i

/kγρ

i

/k

]=− g4nf tr (tart

brtdrtcr)

∫ddk

(2π)d1

(k2)4

[4(gµνgρσ + gµρgνσ − gµσgνρ)(k2)2

− 8(gµνkρkσ + gµρkνkσ + gνσkµkρ + gρσkµkν)k2 + 32kµkνkρkσ]

=− g4nf tr (tartbrtdrtcr)

∫ddk

(2π)d1

(k2)2

[4(gµνgρσ + gµρgνσ − gµσgνρ)

− 4(gµνgρσ + gµρgνσ) + 43

(gµνgρσ + gµρgνσ + gµσgνρ)]

=− 4ig4nf3(4π)2

2

εtr (tart

brtdrtcr)(g

µνgρσ + gµρgνσ − 2gµσgνρ). (16.45)

Combined with the similar diagrams with different permutations, we get

− 8ig4nf3(4π)2

2

ε

[tr (tart

brtcrtdr)(g

µνgρσ − 2gµρgνσ + gµσgνρ)

+ tr (tartbrtdrtcr)(g

µνgρσ + gµρgνσ − 2gµσgνρ)

+ tr (tartcrtbrtdr)(−2gµνgρσ + gµρgνσ + gµσgνρ). (16.46)


Now, we sum up the first four types of diagrams, namely, (16.38), (16.40), (16.42), and

(16.44), and find the result to be

2ig4

3(4π)2

2

ε

[tr (tatbtctd)(−gµνgρσ + 2gµρgνσ − gµσgνρ)

+ tr (tatbtdtc)(−gµνgρσ − gµρgνσ + 2gµσgνρ)

+ tr (tatctbtd)(2gµνgρσ − gµρgνσ − gµσgνρ)]

=2ig4

3(4π)2

2

ε

[gµνgρσ

(2 tr (tatctbtd)− tr (tatbtctd)− tr (tatbtdtc)

)+ gµρgνσ

(2 tr (tatbtctd)− tr (tatbtdtc)− tr (tatctbtd)

)+ gµσgνρ

(2 tr (tatbtdtc)− tr (tatbtctd)− tr (tatctbtd)

)]=

ig4

3(4π)2

2

εC2(G)

[gµνgρσ(−fadef bce − facef bde)

+ gµρgνσ(fadef bce − fabef cde) + gµσgνρ(facef bde + fabef cde)]

=− ig4

3(4π)2

2

εC2(G)

[fabef cde(gµρgνσ − gµσgνρ)

+ facef bde(gµνgρσ − gµσgνρ) + fadef bce(gµνgρσ − gµρgνσ)]. (16.47)

Similar manipulations on (16.46) gives

− 4ig4

3(4π)2

2

εnfC2(r)

[fabef cde(gµρgνσ − gµσgνρ)

+ facef bde(gµνgρσ − gµσgνρ) + fadef bce(gµνgρσ − gµρgνσ)]. (16.48)

Therefore, we finally find δ4g1 to be

− g2

3(4π)2

[C2(G) + 4nfC2(r)

]( 2

ε− logM2

), (16.49)

and it is straightforward to see that δ4g1 − δ3 = 2(δ1 − δ2).

Chapter 17

Quantum Chromodynamics

17.1 Two-Loop renormalization group relations

(a) In this problem we study the higher orders of QCD β function. Formally, we have

β(g) = − b0

(4π)2g3 − b1

(4π)4g5 − b2

(4π)6g7 + · · · . (17.1)

The we can deduce the corresponding function for αs ≡ g2/(4π), namely,

µ∂αs∂µ

= − 2b0

4πα2s −

2b1

(4π)2α3s −

2b2

(4π)3α4s + · · · . (17.2)

Integrate this equation, we get∫ Q

Λ

dµ

2µ= −

∫ αs(Q2)

∞dαs

[b0

4πα2s +

b1

(4π)2α3s +

b2

(4π)3α4s + · · ·

]−1

. (17.3)

The integral can be carried out approximately, as

log(Q/Λ)2 =4π

b0

[1

αs(Q2)+

b1

4πb0

logαs(Q

2)

1 + b14πb0

αs(Q2)+ · · ·

]. (17.4)

Then the running coupling αs(Q2) can be solved iteratively, to be,

αs(Q2) =

4π

b0

[1

log(Q/Λ)2− b1

b20

log log(Q/Λ)2

[log(Q/Λ)2]2+ · · ·

]. (17.5)

(b) Now we substitute (17.5) into the e+e− annihilation cross section, we get

σ(e+e− → hadrons)

= σ0 ·(

3∑f

Q2f

)·[1 +

αsπ

+ a2

( αsπ

)2

+O(α3s)

]= σ0 ·

(3∑f

Q2f

)·[1 +

4

b0

1

log(Q/Λ)2− 4b1

b30

log log(Q/Λ)2

[log(Q/Λ)2]2+ · · ·

]. (17.6)

Since the expression for the cross section is independent of renormalization scheme to the or-

der showed above, we conclude that the β function coefficients b0 and b1 are also independent

of the renormalization scheme.

133

134 Chapter 17. Quantum Chromodynamics

17.2 A Direct test of the spin of the gluon

(a) We repeat the calculations in Part (c) of the Final Project I, with the gluon-quark

vertex replaced by a Yukawa vertex.

iM = Qq(−ie)2(−ig)u(k1)

[i

/k1 + /k3

γµ − γµ i

/k2 + /k3

]v(k2)

−i

q2v(p2)γµu(p1) (17.7)

Then, use the trick described in Final Project I, we have

1

4

∑|iM|2 =

Q2qg

2e4

4s2tr (γµ/p1

γρ/p2)

× tr

[( 1

/k1 + /k3

γµ − γµ 1

/k2 + /k3

)/k2

(γρ

1

/k1 + /k3

− 1

/k2 + /k3

γρ)/k1

]=

32Q2qg

2e4

3s2(p1 · p2)(k1 · k3)(k2 · k3)

[1

(k1 + k3)2+

1

(k2 + k3)2

]2

. (17.8)

Rewrite this in terms of xq, xq and x3, we get

1

4

∑|iM|2 =

4Q2qg

2e4

3s2(1− xq)(1− xq)

[1

1− xq+

1

1− xq

]2

=4Q2

qg2e4

3s2

x23

(1− xq)(1− xq). (17.9)

Note the phase space integral for 3-body final state is deduced in Final Project 1 to be∫dΠ3 =

s

128π3

∫dxqdxq,

thus the differential cross section is given by

d2σ

dx1dx2

(e+e− → qqS) =s

128π3· 1

4|M|2 =

4πα2Q2q

3s· αg

4π

x23

(1− xq)(1− xq). (17.10)

(b) Now let xa > xb > xc. Then there are six ways to associated the original three

variables xq, xq and x3 to these three ordered ones. Note that the integral measure dxadxbdoes not change for different possibilities since the change of integral variables (xq, xq) →(xq, x3) or → (xq, x3) generate an Jacobian whose absolute value is 1, due to the constraint

xq + xq + x3 = 2. Therefore, summing up all 6 possibilities, we get

d2σ

dxadxb(e+e− → qqS)

∝ x2c

(1− xa)(1− xb)+

x2b

(1− xc)(1− xa)+

x2a

(1− xb)(1− xc), (17.11)

for qqS final state, and

d2σ

dxadxb(e+e− → qqS)

∝ x2a + x2

b

(1− xa)(1− xb)+

x2b + x2

c

(1− xb)(1− xc)+

x2c + x2

a

(1− xc)(1− xa), (17.12)

We plot these two distributions on the xa − xb plain with the range xa > xb > xc, as shown

in Figure

17.3. Quark-gluon and gluon-gluon scattering 135

0.6 0.7 0.8 0.9 1.0

0.0

0.2

0.4

0.6

0.8

1.0

xa

xb

0 1

0.6 0.7 0.8 0.9 1.0

0.0

0.2

0.4

0.6

0.8

1.0

xa

xb

0 1

Figure 17.1: The differential cross sections of e+e− → qqg as a function of x1 and x2,

assuming gluon is a vector/scalar particle in left/right diagram.

17.3 Quark-gluon and gluon-gluon scattering

In this problem we evaluate the cross sections for two processes: (a) qq → gg, (b)

gg → gg.

(a) There are three diagrams contributing the process q(k1)q(k2)→ g(p1)g(p2) at the tree

level, as shown in Fig. 17.11 of Peskin&Schroeder. The amplitudes associated with these

diagrams are listed as follows:

iM1 = (ig)2v(k2)/ε∗(p2)i(/k1 − /p1

)

(k1 − p1)2/ε∗(p1)u(k1)tbta, (17.13a)

iM2 = (ig)2v(k2)/ε∗(p1)i(/k1 − /p2

)

(k1 − p2)2/ε∗(p2)u(k1)tatb, (17.13b)

iM3 = (ig)gfabc[gµν(p2 − p1)ρ − gνρ(2p2 + p1)µ + gρµ(p2 + 2p1)ν

]× −i

(k1 + k2)2v(k2)γρu(k1)ε∗µ(p1)ε∗ν(p2)tc. (17.13c)

It is convenient to evaluate these diagrams with initial and final states of definite helicities.

By P and CP symmetry of QCD, there are only two independent processes, namely qLqR →gRgR and qLqR → gRgL, that could be nonzero. Let’s evaluate them in turn for the three

diagrams. To begin with, we set up the kinematics:

kµ1 = (E, 0, 0, E), pµ1 = (E,E sin θ, 0, E cos θ),

kµ2 = (E, 0, 0,−E), pµ2 = (E,−E sin θ, 0,−E cos θ). (17.14)

Then,

uL(k1) =√

2E(0, 1, 0, 0), vL(k2) =√

2E(1, 0, 0, 0).


ε∗Lµ(p1) = 1√2(0,− cos θ,−i, sin θ), ε∗Rµ(p1) = 1√

2(0,− cos θ, i, sin θ),

ε∗Lµ(p2) = 1√2(0, cos θ,−i,− sin θ), ε∗Rµ(p2) = 1√

2(0, cos θ, i,− sin θ). (17.15)

Now we begin the calculation. (In the following we use sθ ≡ sin θ and cθ = cos θ.)

iM1(qLqR → gRgR) =−ig2E2tbta

t(0, 0, 1, 0)

−sθ 1 + cθ−1 + cθ sθ

sθ −1− cθ1− cθ −sθ

×

1− cθ −sθ−sθ −1 + cθ

−1 + cθ sθsθ 1− cθ

sθ 1− cθ−1− cθ −sθ

−sθ −1 + cθ1 + cθ sθ

0

1

0

0

= ig2tbta

2E2

t(1− cos θ) sin θ = −ig2tbta sin θ. (17.16)

iM2(qLqR → gRgR) =−ig2E2tatb

u(0, 0, 1, 0)

sθ 1− cθ

−1− cθ −sθ−sθ −1 + cθ

1 + cθ sθ

×

1 + cθ sθsθ −1− cθ

−1− cθ −sθ−sθ 1 + cθ

−sθ 1 + cθ−1 + cθ sθ

sθ −1− cθ1− cθ −sθ

0

1

0

0

= −ig2tbta

2E2

u(1 + cos θ) sin θ = ig2tatb sin θ. (17.17)

iM3(qLqR → gRgR) =g2fabctcE2

s(0, 0, 1, 0)

−4

cθ sθsθ −cθ

−cθ −sθ−sθ cθ

0

1

0

0

= g2fabctc sin θ = −ig2[ta, tb] sin θ. (17.18)

Thus we find that

iM(qLqR → gRgR) =(iM1 + iM2 + iM3

)(qLqR → gRgR) = 0. (17.19)

In the same manner, we calculate the amplitude for qLqR → gRgL. This time, we find:

iM1(qLqR → gRgL) =− ig2tbta sin θ, (17.20a)

iM2(qLqR → gRgL) =− ig2tatbt

usin θ, (17.20b)

17.3. Quark-gluon and gluon-gluon scattering 137

iM3(qLqR → gRgL) = 0, (17.20c)

Therefore,

iM(qLqR → gRgL) = −ig2(tbta + tatb

t

u

)sin θ, (17.21)

and by crossing symmetry,

iM(qLqR → gLgR) = −ig2(tbta + tatb

u

t

)sin θ. (17.22)

There are two more nonzero amplitudes with qRqL initial states, which are identical to the

amplitudes above. Then we find the spin- and color-summed/averaged squared amplitude

to be

1

32· 1

22

∑spin,color

|M|2

=1

36· 2 · g4 sin2 θ

[(tr (tbtatatb) + 2 tr (tbtatbta)

t

u+ tr (tatbtbta)

t2

u2

)+ (t↔ u)

]=

8π2α2s

9(1− cos2 θ)

[(16

3

(1 +

t2

u2

)− 4t

3u

)+ (t↔ u)

]=

512π2α2s

27

[t

u+u

t− 9(t2 + u2)

4s2

]. (17.23)

Therefore the differential cross section is given by

dσ

dt=

32πα2s

27s2

[t

u+u

t− 9(t2 + u2)

4s2

]. (17.24)

(b) Now consider the process g(k1)g(k2) → g(p1)g(p2). The four tree level diagrams are

shown in Fig. 17.12 of Peskin&Schroeder. Their amplitudes are given by:

iM1 = g2fabcf cde−i

s

[gµν(k1 − k2)λ + gνλ(k1 + 2k2)µ − gλµ(2k1 + k2)ν

]×[gρσ(p2 − p1)λ − gσλ(p1 + 2p2)ρ + gρλ(p1 + 2p2)ρ

]εµ(k1)εν(k2)ε∗ρ(p1)ε∗σ(p2),

(17.25a)

iM2 = g2facef bde−i

t

[gµρ(k1 + p1)λ − gρλ(2p1 − k1)µ − gλν(2k1 − p1)ρ

]×[gνσ(k2 + p2)λ − gσλ(2p2 − k2)ν + gνλ(p2 − 2k2)σ


(17.25b)

iM3 = g2fadef bce−i

u

[gµσ(k1 + p2)λ − gσλ(2p2 − k1)µ − gλν(2k1 − p2)σ

]×[gνρ(k2 + p1)λ − gρλ(2p1 − k2)ν + gνλ(p1 − 2k2)ρ


(17.25c)

iM4 =− ig2[fabef cde

(ε(k1) · ε∗(p1)ε(k2) · ε∗(p2)− ε(k1) · ε∗(p2)ε(k2) · ε∗(p1)

)+ facef bde

(ε(k1) · ε(k2)ε∗(k1) · ε∗(p2)− ε(k1) · ε∗(p2)ε(k2) · ε∗(p1)

)+ fadef bce

(ε(k1) · ε(k2)ε∗(p1) · ε∗(p2)− ε(k1) · ε∗(p1)ε(k2) · ε∗(p2)

)]. (17.25d)


The choice for all external momenta and final states polarizations are the same with that in

(a). Now to evaluate the amplitude gRgR → gRgR, we also need the initial states polarization

vectors for right-handed gluons with momenta k1 and k2, which are given by

εµR(k1) = 1√2(0, 1, i, 0), εµR(k2) = 1√

2(0,−1, ı, 0). (17.26)

Then after some calculations, we find,

iM1 =− ig2fabef cde cos θ, (17.27a)

iM2 = ig2facef bde19 + 7 cos θ − 11 cos2 θ + cos3 θ

4(1− cos θ); (17.27b)

iM3 = ig2fadef bce19− 7 cos θ − 11 cos2 θ − cos3 θ

4(1 + cos θ); (17.27c)

iM4 =− ig2[fabef cde cos θ + 1

4facef bde(3 + 2 cos θ − cos2 θ)

+ 14fadef bce(3− 2 cos θ − cos2)

]. (17.27d)

The sum of these four amplitudes is

iM(gRgR → gRgR) =− 2ig2[fabef cde cos θ − facef bde

( 2

1− cos θ+ cos θ

)− fadef bce

( 2

1 + cos θ− cos θ

)]= 4ig2

[facef bde

1

1− cos θ+ fadef bce

1

1 + cos θ

]=− 2ig2

[facef bde

s

t+ fadef bce

s

u

]. (17.28)

We can also obtain the amplitudes for gLgR → gLgR and gLgR → gRgL from the result above

by crossing symmetry, namely the change of variables (s, b) ↔ (u, d) and (s, b) ↔ (t, c),

which gives

iM(gLgR → gLgR) = 2ig2[facef bde

u

t+ fabef cde

u

s

], (17.29)

iM(gLgR → gRgL) =− 2ig2[fabef cde

t

s− fadef bce t

u

]. (17.30)

The amplitudes for gLgL → gLgL, gRgL → gRgL and gRgL → gLgR are identical to the

amplitudes for gRgR → gRgR, gLgR → gLgR and gLgR → gRgL, respectively, due to parity

conservation of QCD. It can be shown by the conservation of angular momentum that other

helicity amplitudes all vanish. Therefore we have found all required amplitude. To get the

cross section, we take the square of these results.∑|M(gRgR → gRgR)|2

= 4g4[facef bdefacff bdf

s2

t2+ fadef bcefadff bcf

s2

u2+ 2facef bdefadff bcf

s2

tu

]= 4g4

[tr (tatatbtb)

( s2

t2+s2

u2

)+ 2 tr (tatbtatb)

s2

tu

]

17.4. The gluon splitting function 139

= 288g4( s2

t2+s2

u2+s2

tu

), (17.31)

where ta is the generator of SU(3) group in adjoint representation which is related to the

structure constants by fabc = i(ta)bc. Thus tr (tatatbtb) = (C2(G))2d(G) = 72, and

tr (tatbtatb) = tr (tatb[ta, tb]) + tr (tatatbtb) = 12

ifabc tr ([ta, tb]tc) + (C2(G))2d(G)

=− 12fabcfabd tr (tctd) + (C2(G))2d(G) = 1

2(C2(G))2d(G),

which is 36 for SU(3). Similarly, we can work out the square of other amplitudes, to be∑|M(gLgR → gLgR)|2 = 288g4

( u2

t2+u2

s2+u2

st

), (17.32)∑

|M(gLgR → gRgL)|2 = 288g4( t2s2

+t2

u2+t2

su

). (17.33)

Therefore, the spin-averaged and squared amplitudes is

1

82· 1

22

∑|M2| = 1

82 · 22· 2 · 288g4

(6− 2tu

s2− 2us

t2− 2st

u2

)= 72π2α2

s

(3− tu

s2− us

t2− st

u2

). (17.34)

Thus the differential cross section is

dσ

dt(gg → gg) =

9πα2s

2s2

(3− tu

s2− us

t2− st

u2

). (17.35)

17.4 The gluon splitting function

In this problem we calculate the gluon splitting function Pg←g(z) by evaluating the

amplitude of the virtual process g → gg, as shown in Fig. 17.2.

Figure 17.2: The Gluon splitting process.

The momenta of initial and final states are taken to be the same with that of Fig. 17.16

of Peskin&Schroeder. That is, we have

p = (p, 0, 0, p), q = (zp, p⊥, 0, zp), k = ((1− z)p,−p⊥, 0, (1− z)p), (17.36)

and the polarization vectors associated with gluons are,

εiL(p) = 1√2(1,−i, 0), εiR(p) = 1√

2(1, i, 0),

εiL(q) = 1√2(1,−i,− p⊥

zp), εiR(q) = 1√

2(1, i,− p⊥

zp),


εiL(k) = 1√2(1,−i,

p⊥(1− z)p

), εiR(k) = 1√2(1, i,

p⊥(1− z)p

). (17.37)

Then we can evaluate the amplitude for the process g → gg directly, which is given by

iMabc = gfabc[(ε∗(q) · ε(p)

)((p+ q) · ε∗(k)

)+(ε∗(q) · ε∗(k)

)((k − q) · ε(p)

)−(ε∗(k) · ε(p)

)((p+ k) · ε∗(q)

)]. (17.38)

We evaluate the amplitudes with definite initial and final polarizations in turn:

iMabc(gL(p)→ gL(q)gL(k)) =√

2( 1

1− z+

1

z

)gfabcp⊥, (17.39a)

iMabc(gL(p)→ gL(q)gR(k)) =

√2z

1− zgfabcp⊥, (17.39b)

iMabc(gL(p)→ gR(q)gL(k)) =

√2(1− z)

zgfabcp⊥, (17.39c)

iMabc(gL(p)→ gR(q)gR(k)) = 0. (17.39d)

By parity invariance, the amplitudes with right-handed initial gluon are dictated by the

results above. Note further that fabcfabc = 24, thus we have

12· 1

8

∑spin,color

|M|2 = 12· 1

8· 2 · 24 · 2g2p2

⊥

×[( 1

1− z+

1

z

)2

+z2

(1− z)2+

(1− z)2

z2

]=

12g2p2⊥

z(1− z)

[ 1− zz

+z

1− z+ z(1− z)

]=

2e2p2⊥

z(1− z)· P (0)

g←g(z), (17.40)

where the superscript (1) represents the part of the splitting function contributed from the

diagram calculated above, in parallel with the notation of Peskin&Schroeder. (See 17.100.

for instance.) Therefore we get

P (1)g←g = 6

[ 1− zz

+z

1− z+ z(1− z)

]. (17.41)

Besides, there should be a term proportional to δ(1−z) in Pg←g, which comes from the zeroth

order, as well as the corrections from Pq←g and Pg←g, where Pq←g(z) = 12

(z2 + (1 − z)2).

Now let’s take it to be Aδ(1− z), then the coefficient A can be determined by the following

normalization condition (namely the momentum conservation):

1 =

∫ 1

0

dz z[2nfPq←g(z) + P (1)

g←g(z) + Aδ(1− z)], (17.42)

where nf is the number of fermion types, and the coefficient 2 is from contributions of both

quarks and anti-quarks. To carry out the integral, we use the prescription 11−z →

1(1−z)+ ,

then it is straightforward to find that A = 112− 1

3nf . Therefore,

Pg←g = 6[ 1− z

z+

z

(1− z)+

+ z(1− z)]

+( 11

2− nf

3

)δ(1− z). (17.43)

17.5. Photoproduction of heavy quarks 141

17.5 Photoproduction of heavy quarks

In this problem we study the production of a pair of heavy quark-antiquark by the

scattering of a photon off a proton. At the leading order at the parton level, the process is

contributed from the photon-gluon scattering, as shown in Figure 17.3.

Figure 17.3: Tree diagrams for the photoproduction of heavy quarks at the parton level.

The corresponding amplitude can be read from a similar process γγ → e+e− in QED.

From (5.105) of Peskin & Schroeder, we have the amplitude for e+e− → 2γ, which reads

(adapted to our notation for external momenta)

1

4

∑|M(e+e− → 2γ)|2 = 2e4

[k1 · p2

k1 · p1

+k1 · p1

k1 · p2

+ 2m2( 1

k1 · p1

+1

k1 · p2

)−m4

( 1

k1 · p1

+1

k1 · p2

)2]. (17.44)

Then the amplitude M(γg → QQ) can be obtained by making the exchange (k1, k2) ↔(p1, p2), replacing e4 by e2g2, and also including the factor 1

8Q2q tr (tata) = 1

2Q2q taking

account of the color average, the electric charge of quarks, and the summation of color

indices, respectively. Then the amplitude in the present case is

1

4 · 8∑|M(γg → QQ)|2 = e2g2Q2

q

[p1 · k2

p1 · k1

+p1 · k1

p1 · k2

+ 2m2( 1

p1 · k1

+1

p1 · k2

)−m4

( 1

p1 · k1

+1

p1 · k2

)2]. (17.45)

In parton’s center-of-mass frame, we have k1 = (E, 0, 0, E), k2 = (E, 0, 0,−E), p1 =

(E, p sin θ, 0, p cos θ) and p2 = (E,−p sin θ, 0,−p cos θ), with p2 = E2 −m2. Then p1 · k1 =

E(E − p cos θ) and p1 · k2 = E(E + p cos θ). Then the differential cross section is

dσ

d cos θ=

πααsQ2q

16

p

E3

[E2 + p2 cos2 θ − 2m2

E2 − p2 cos2 θ− 2m4

(E2 − p2 cos2 θ)2

]. (17.46)

Then the cross section for photon and proton initial state is given by

σ(γ(k1) + p(k2)→ QQ) =

∫dx fg(x)σ(γ(k1) + g(xk2)→ QQ). (17.47)


17.6 Behavior of parton distribution functions at small

x

(a) In this problem we study the solution of A-P equations at small x with certain ap-

proximations. Firstly, we show that the A-P equations,

d

d logQfg(x,Q) =

αs(Q2)

π

∫ 1

x

dz

z

[Pg←q(z)

∑f

(ff

( xz,Q)

+ ff

( xz,Q))

+ Pg←g(z)fg

( xz,Q)], (17.48)

d

d logQff (x,Q) =

αs(Q2)

π

∫ 1

x

dz

z

[Pq←q(z)ff

( xz,Q)

+ Pq←g(z)fg

( xz,Q)], (17.49)

d

d logQff (x,Q) =

αs(Q2)

π

∫ 1

x

dz

z

[Pq←q(z)ff

( xz,Q)

+ Pq←g(z)fg

( xz,Q)], (17.50)

can be rewritten as a differential equation with variable ξ = log log(Q2/Λ2). To see this,

we note that d/d logQ = 2e−ξd/dξ, and to 1-loop order, αs(Q) = 2π/(b0 log(Q/Λ)

)=

(4π/b0)e−ξ, so we have

d

dξfg(x, ξ) =

2

b0

∫ 1

x

dz

z

[Pg←q(z)

∑f

(ff

( xz, ξ)

+ ff

( xz, ξ))

+ Pg←g(z)fg

( xz, ξ)], (17.51)

d

dξff (x, ξ) =

2

b0

∫ 1

x

dz

z

[Pq←q(z)ff

( xz, ξ)

+ Pq←g(z)fg

( xz, ξ)], (17.52)

d

dξff (x, ξ) =

2

b0

∫ 1

x

dz

z

[Pq←q(z)ff

( xz, ξ)

+ Pq←g(z)fg

( xz, ξ)]. (17.53)

(b) Now we apply the approximation that 1) gluon PDF dominates the integrand in the

A-P equations and 2) the function g(x,Q) = xfg(x,Q) is slowly varying with x when x is

small. Then, define w = log(1/x), which gives d/dw = −xd/dx, we can calculate

∂2

∂w∂ξg(x, ξ) =− x d

dx

(x∂

∂ξfg(x,Q)

)'− x d

dx

(2x

b0

∫ 1

x

dz

zPg←g(z)fg

( xz,Q))

=2

b0

· xPg←g(x)fg(x,Q)− 2x

b0

∫ 1

x

dzPg←g(z)d

dx

[ xzfg

( xz,Q)]

' 2

b0

· xPg←g(x)fg(x,Q). (17.54)

From the result of Problem 17.4 we know that xPg←g(x) = 6 as x→ 0. Therefore the A-P

equation for fg becomes

∂2

∂w∂ξg(x, ξ) =

12

b0

g(x, ξ). (17.55)

17.6. Behavior of parton distribution functions at small x 143

Then we verify that

g = K(Q2) · exp

([ 48

b0

w(ξ − ξ0)]1/2)

(17.56)

is an approximation solution to the differential equation above when wξ 1, where K(Q2)

is an initial condition. We apply ∂2/∂w∂ξ on this expression, to get

∂2

∂w∂ξg(w, ξ) =

1

4

√48

b0w(ξ − ξ0)· exp

([ 48

b0

w(ξ − ξ0)]1/2)

×[2(ξ − ξ0)

∂K(Q2)

∂ξ+

(1 +

√48

b0

w(ξ − ξ0)

)K(Q2)

]. (17.57)

In the limit wξ 1, the square root term in the last line dominates, thus

∂2

∂w∂ξg(w, ξ) ' b0

12K(Q2) exp

([ 48

b0

w(ξ − ξ0)]1/2)

=12

b0

g(w, ξ). (17.58)

(c) Then we consider the A-P equation for quarks. If we adopt the approximation in (b)

again, namely, the gluon PDF dominates and the function q(x, ξ) = xff (x,Q) is slowly

varying, then we have

∂

∂ξq(x, ξ) = x

∂

∂ξff (x, ξ) =

2x

b0

∫ 1

x

dz

zPq←g(z)fg

( xz, ξ)

=2

b0

∫ 1

x

dzPq←g(z)g( xz, ξ)

=1

b0

∫ 1

x

dz(z2 + (1− z)2

)g( xz, ξ)

' 2

b0

[1

6(2z3 − 2z2 + 3z)g

( xz, ξ)]1

x

' 2

3b0

g(x, ξ), (17.59)

where we have used x 1 and ∂q(x, ξ)/∂x ' 0. Then, we verify that

q =

√ξ − ξ0

27bowK(Q2) · exp

([ 48

b0

w(ξ − ξ0)]1/2)

(17.60)

is again an approximate solution to the equation derived above, in the limit wξ 1. In

fact,

∂

∂ξq(x, ξ) = exp

([ 48

b0

w(ξ − ξ0)]1/2)[

2

3b0

K(Q2)

+1

18

√3

b0w(ξ − ξ0)

(K(Q2) + 2(ξ − ξ0)

∂K(Q2)

∂ξ

)]' 2

3b0

K(Q2) exp

([ 48

b0

w(ξ − ξ0)]1/2)

=2

3b0

g(x, ξ). (17.61)

(d) We use the fitted formula of K(Q2) to plot the PDFs of gluon and quarks in Figure.


0.00 0.05 0.10 0.15 0.200.0

0.5

1.0

1.5

2.0

2.5

3.0

x

xfgHx,Q2L

xf f Hx,Q2L

Figure 17.4: Approximate parton distribution functions at small x with Q = 500GeV.

Chapter 18

Operator Products and Effective

Vertices

18.1 Matrix element for proton decay

(a) We estimate the order of magnitude of the proton lifetime, through the decay p →e+π0, based on the following operator,

OX =2

m2X

εijkεαβεγδeRαuRiβuLjγdLkδ, (18.1)

where mX is the scale of this higher dimensional operator, whose typical value is around

1016GeV, and i, j, k are color indices for quarks, α, β, · · · are spinor indices. Then, the

amplitudeM of this decay process should be proportional to m−2X . Note that the amplitude

M has mass dimension 1, thus we should haveM∼ m3pm−2X with mp the proton mass. Now

take mX ∼ 1016GeV and mp ∼ 1GeV, we have the decay width

Γ ∼ 1

8π

1

2mp

|M|2 ∼ 1

16π

m5p

m4X

∼ 10−65GeV ∼ 1033yr−1. (18.2)

(b) Now we consider the first order QCD correction to the estimation above. The correc-

tion comes from virtual gluon exchange among three quarks in the operator. To evaluate

these 1-loop diagrams, we firstly fixed the renormalization condition of OX to be

uLjγ dLkδ

uRiβ

α= iεijkδ

αβεγδ. (18.3)

The 1-loop diagrams are shown in Figure 18.1. The Feynman rules can be written in two-

component spinor notations. The left-handed spinor’s propagator reads i(p ·σ)/p2, the right-

handed spionr’s propagator is i(p · σ)/p2, the QCD interaction between quark and gluon is

i[ψ†Liσµ(ta)ijψLj +ψ†Riσ

µ(ta)ijψRj], and the vertex corresponding OX reads iεijkδαβεγδ. Then

145

146 Chapter 18. Operator Products and Effective Vertices

Figure 18.1: 1-loop QCD correction to the effective operator of proton decay.

the first diagram reads

(a) = i(ig)2εimn(ta)mj(ta)nkδ

αβεγ′δ′∫

ddq

(2π)d−i

q2

(−iq · σq2

σµ

)γ′γ

(iq · σq2

σµ)δ′δ

=− g2 ·(− 2

3

)εijkδ

αβεγ′δ′(σρσµ)γ′

γ(σσσµ)δ′

δ

∫ddq

(2π)dqρqσ

q6

=− g2 ·(− 2

3

)εijkδ

αβ · 16εγδ · i

4(4π)2

2

ε

=8g2

3(4π)2

2

ε· iεijkδαβεγδ, (18.4)

where the Pauli matrices is simplified as follows,

εγ′δ′(σρσµ)γ′

γ(σσσµ)δ′

δ = (σTµσTρ εσ

ρσµ)γδ = −(σTµ εσρσρσµ)γδ

= −4(σTµ εσµ)γδ = −4(εσµσ

µ)γδ = 16εγδ, (18.5)

in which we used the fact that εαβ = iσ2 and σTµσ2 = −σ2σµ. In the computation of this

diagram, we also used εimn(ta)mj(ta)nk = −(2/3)εijk. The coefficient of this equality can

be easily justified by contracting both sides with εijk. Similarly, we compute the second

diagram, as follows,

(b) = i(ig)2εimn(ta)mj(ta)nkδ

αβ′εγ′δ

∫ddq

(2π)d−i

q2

(−iq · σq2

σµ

)β′β

(iq · σq2

σµ)γ′γ

=− g2 ·(− 2

3

)εijk · 4δαβεγδ ·

i

4(4π)2

2

ε

=2g2

3(4π)2

2

ε· iεijkδαβεγδ, (18.6)

where we used the identity (σµ)αβ(σµ)γδ = 2δαδδβγ. The third diagram gives the same result

as the second one. Therefore, we get the counterterm for the operator OX in MS scheme

to be

δOX = − 4g2

(4π)2

(2

ε− logM2

), (18.7)

where M2 is the renormalization scale. We further recall that the field strength renormal-

ization counterterm for quarks in QCD is given by

δ2 = − 4g2

3(4π)2

(2

ε− logM2

), (18.8)

18.2. Parity-violating deep inelastic form factor 147

then the anomalous dimension of operator OX is given by

γ = M∂

∂M

(− δOX +

3

2δ2

)= − 4g2

(4π)2. (18.9)

Therefore this QCD correction will enhance the operator strength by a factor of(log(m2

X/Λ2)

log(m2p/Λ

2)

)a0/2b0, (18.10)

where Λ ' 200GeV, a0 = 4 is the coefficient from anomalous dimension, and b0 = 11 −(2/3)nf = 7 is the 1-loop coefficient of QCD β function. Taking mX = 1016GeV and

mp = 1GeV, this factor is about 2.5. Then the decay rate of proton is enhanced by a factor

of 2.52 ' 6.3.

18.2 Parity-violating deep inelastic form factor

(a) We firstly compute the amplitude of the neutrino deep inelastic scattering through

charged current interaction, which reads

iM(νp→ µ−X) =ig2

2m2W

u(k′)γµ

( 1− γ5

2

)u(k)

×∫

d4x eiq·x〈X|(Jµ+(x) + Jµ−(x)

)|P 〉. (18.11)

Then the squared amplitude with initial proton’s spins averaged and final state X summed

is

1

2

∑|M|2 =

1

2

g4

4m4W

∑spin

(u(k′)γν

( 1− γ5

2

)u(k)u(k)

( 1 + γ5

2

)γµu(k′)

)×∑X

∫dΠX〈P |

(Jµ+(x) + Jµ−(x)

)|X〉〈X|

(Jν+(0) + Jν−(0)

)|P 〉. (18.12)

The trace factor can be straightforwardly worked out to be

Lµν ≡∑spin

(u(k′)γµ

( 1− γ5

2

)u(k)u(k)

( 1 + γ5

2

)γνu(k′)

)= tr

[γµ

( 1− γ5

2

)/k( 1 + γ5

2

)γν/k

′]

= 2(kµk

′ν + kνk

′µ − gµνk · k′ + iεµν

ρσkρk′σ

). (18.13)

Then, use the optical theorem, we have

Lµν∑X

∫dΠX〈P |

(Jµ+(x) + Jµ−(x)

)|X〉〈X|

(Jν+(0) + Jν−(0)

)|P 〉

= 2 Im(LµνW

µν(ν)), (18.14)


with

W µν(ν) = 2i

∫d4x eiq·x〈P |T

Jµ−(x)Jν+(0)

|P 〉, (18.15)

Therefore, the cross section is

σ(νp→ µ−X) =1

2s

∫d3k′

(2π)3

1

2k′· 1

2

∑|M|2

=1

2s

∫dxdy

ys

(4π)2· g4

4m4W

Im(LµνW

µν(ν)), (18.16)

and the differential cross section is thus given by

d2σ

dxdy(νp→ µ−X) =

yG2F

2π2Im[(kµk

′ν + kνk


ρσkρk′σ

)W µν(ν)

]. (18.17)

(b) The lepton momentum tensor obtained in (a) is

Lµν = 2(kµk′ν + kνk


ρσkρk′σ). (18.18)

Then it is straightforward to see that qµLµν = (k − k′)µLµν = 0 and qνLµν = 0. As a

consequence, any term in W µν(ν) proportional to qµ or qν is irrelevant. Therefore we can

rewrite the tensor W µν(ν) in terms of three form factors W(ν)i (i = 1, 2, 3). That is,

W µν(ν) = −gµνW (ν)1 + P µP νW

(ν)2 + iεµνρσPρqσW

(ν)3 + · · · . (18.19)

Then the deep inelastic scattering cross section becomes

d2σ

dxdy(νp→ µ−X) =

yG2F

2π2

[2(k · k′) ImW

(ν)1 + 2(P · k)(P · k′) ImW

(ν)2

− 4(

(P · k)(q · k′)− (q · k)(P · k′))

ImW(ν)3

]. (18.20)

(c) Now we evaluate ImW(ν)1,2,3 in the parton model. Firstly, W µν(ν) can be written as

W µν(ν) = 2i

∫d4x eiq·x

∫ 1

0

dξ∑f

ff (ξ)1

ξ

⟨qf (ξP )

∣∣TJµ−(x)Jν+(0)∣∣qf (ξP )

⟩, (18.21)

and be evaluated in terms of Feynman diagrams displayed in Fig. 18.10 of Peskin & Schroeder.

For the first diagram, we have

2i

∫ 1

0

dξ

[fd(ξ)

1

ξu(p)γµ

( 1− γ5

2

) i

/p+ /q + iεγν( 1− γ5

2

)u(p)

+fu(ξ)1

ξu(p)γν

( 1− γ5

2

) i

/p+ /q + iεγµ( 1− γ5

2

)u(p)

], (18.22)

where p = ξP . Then, averaging/summing over initial/final spin states gives

2

∫ 1

0

dξ

[fd(ξ)

1

ξ· 1

2tr(/pγ

µ 1− γ5

2(/p+ /q)γ

ν 1− γ5

2

)

18.2. Parity-violating deep inelastic form factor 149

+ fu(ξ)1

ξ· 1

2tr(/pγ

µ 1− γ5

2(/p+ /q)γ

ν 1− γ5

2

)] −1

2p · q + q2 + iε

⇒∫ 1

0

dξ

ξ

[(fd(ξ) + fu(ξ)

)(4ξ2P µP ν − 2ξP · qgµν

)+(fd(ξ)− fu(ξ)

)2iεµνρσP

ρqσ]

−1

2p · q + q2 + iε, (18.23)

where we have dropped terms containing qµ or qν in the last line. Then it is easy to read

from this expression that

ImW(ν)1 = 2P · q

∫ 1

0

dξ[fd(ξ) + fu(ξ)

]Im( −1

2p · q + q2 + iε

), (18.24)

ImW(ν)2 =

∫ 1

0

dξ 4ξ[fd(ξ) + fu(ξ)

]Im( −1

2p · q + q2 + iε

), (18.25)

ImW(ν)3 = 2

∫ 1

0

dξ[fd(ξ)− fu(ξ)

]Im( −1

2p · q + q2 + iε

), (18.26)

where 2P · q = ys, and

Im( −1

2p · q + q2 + iε

)=

π

ysδ(ξ − x). (18.27)

Note that the second diagram in Fig. 18.10 of Peskin & Schroeder does not contributes, as

explained in the book. Therefore we conclude that

ImW(ν)1 = π

[fd(x) + fu(x)

], (18.28)

ImW(ν)2 =

4πx

ys

[fd(x) + fu(x)

], (18.29)

ImW(ν)3 =

2π

ys

[fd(x)− fu(x)

]. (18.30)

(d) The analysis above can be easily repeated for the left-handed current JµfL of single

flavor f , defined by JµfL = fγµPLf where PL ≡ (1− γ5)/2. Then, define

W µνfL = 2i

∫d4x eiq·x⟨P ∣∣TJµfL(x)JνfL(0)

∣∣P⟩, (18.31)

and its decomposition,

W µνfL = −gµνW1fL + P µP νW2fL + iεµνρσPρqσW3fL + · · · . (18.32)

We see that it amounts to the replacement in the final result that d → f and u → f .

Therefore,

ImW(ν)1fL = π

[ff (x) + ff (x)

], (18.33)

ImW(ν)2fL =

4πx

ys

[ff (x) + ff (x)

], (18.34)

ImW(ν)3fL =

2π

ys

[ff (x)− ff (x)

]. (18.35)


(e) Now we perform OPE on W µνfL . Firstly,∫

d4x eiq·xJµfL(x)JνfL(0) '∫

d4x eiq·x(qγµPLq(x)qγνPLq(0) + qγµPLq(x)qγνPLq(0)

)= qγµPL

i(i/∂ + /q)

(i∂ + q)2γνPLq +

(µ↔ ν, q → −q

). (18.36)

Then, the first term in the last line can be written as

qγµPLi(i/∂ + /q)

(i∂ + q)2γνPLq =

1

2

(qγµ

i(i/∂ + /q)

(i∂ + q)2γνq − qγµ

i(i/∂ + /q)

(i∂ + q)2γνγ5q

)=− i

2q[2γ(µ(i∂ν))− gµν/q − iεµνρσ(i∂ + q)ργσ

]× 1

Q2

∞∑n=0

( 2iq · ∂Q2

)nq, (18.37)

where we have symmetrize the µν indices for the first two terms in the square bracket and

antisymmetrize the indices for the third term, by using the equalities 12

(γµγλγν +γνγλγµ) =

gµλγν +gνλγµ−gµνγλ and 12

(γµγλγνγ5−γνγλγµγ5) = −iεµνλργρ, and terms proportional to

qµ or qν have also been dropped. The (anti)symmetrization can be understood by looking

at (18.32), where the terms with no γ5 are symmetric on µν while the term involving γ5 is

antisymmetric on µν. Therefore, when including the second term in (18.36), we should keep

terms of even powers in q for symmetric µν indices and of odd powers in q for antisymmetric

µν.

Now, with these understood, and using the definition of twist-2, spin-n operator,

O(n)µ1···µnf = qfγ

(µ1(iDµ2) · · · (iDµn))qf − traces, (18.38)

we have,

i

∫d4x eiq·xJµfL(x)JνfL(0) =

∑n>0, even

2(2qµ1) · · · (2qµn−2)

(Q2)n−1O(n)µνµ1···µn−2

f

− 1

2gµν

∑n>0, even

(2qµ1) · · · (2qµn)

(Q2)nO(n)µ1···µnf

− iεµνρσqρ∑

n>0 odd

(2qµ1) · · · (2qµn−1)

(Q2)nO(n)σµ1···µn−1

f . (18.39)

Then, using 〈P |O(n)µ1···µnf |P 〉 = 2AnfP

µ1 · · ·P µn , we can get W µνfL to be

W µνfL = 8P µP ν

∑n>0, even

(2q · P )n−2

(Q2)n−1Anf − 2gµν

∑n>0, even

(2q · P )n

(Q2)nAnf

+ 4iεµνρσPρqσ

∑n>0, odd

(2q · P )n−1

(Q2)nAnf . (18.40)

18.3. Anomalous dimensions of gluon twist-2 operators 151

So we can read out

W1fL = 2∑

n>0, even

(2q · P )n

(Q2)nAnf , (18.41)

W2fL = 8∑

n>0, even

(2q · P )n−2

(Q2)n−1Anf , (18.42)

W3fL = 4∑

n>0, odd

(2q · P )n−1

(Q2)nAnf . (18.43)

(f) Now we use W3fL obtained above to derive a sum rule for parton distribution f−f ,

defined by

f−f (x,Q2) =ys

2πImW3fL(x,Q2), (18.44)

where x = Q2/ν and ν = 2P · q = ys. The the analytic behavior of W3fL on the v-complex

plane is shown in Fig. 18.11 of Peskin & Schroeder. Thus we can define the contour integral

I3n =

∫dν

2πi

1

νnW3fL(ν,Q2), (18.45)

where the contour is a small circle around the origin ν = 0. This integral picks up the

coefficient of ν(n−1) term, namely, I3n = 4Anf/(Q2)n. On the other hand, the contour can be

deformed as shown in Fig. 18.12 of Peskin & Schroeder. Then the integral can be evaluated

as

I3n = 2

∫ ∞Q2f

dν

2πi

1

νn(2i) ImW3fL(ν,Q2) =

4

(Q2)n

∫ 1

0

dx xn−1f−f (x,Q2). (18.46)

Therefore we get the sum rule, ∫ 1

0

dx xn−1f−f (x,Q2) = Anf . (18.47)

18.3 Anomalous dimensions of gluon twist-2 operators

In this problem we finish evaluating anomalous dimension matrix γn in (18.180) of Peskin

& Schroeder, given by

γn = − g2

(4π)2

(anff anfgangf angg

)(18.48)

where anff has already been evaluated explicitly in the book. Here we evaluate the remaining

three elements. The needed Feynman rules involving operators O(n)f and O(n)

g are listed as

follows:

k= /Λ(Λ · k)n−1,


kaµ bν

= −2[gµν(Λ · k)n + k2ΛµΛν(Λ · k)n−2 − 2k(µΛν)(Λ · k)n−1

]δab,

k1 ↑k2

k3

aµbν cλ

=− 2igfabcgµνΛλ

n∑j=1

(Λ · k1)j−1(−Λ · k2)n−j

+ (cyclic permutations on µak1, νbk2, λck3) + · · · .

In the last expression, we list only terms containing a metric tensor gµν , and the ignored

terms (marked by · · · ) are irrelevant in the following calculations. To be clear, we have

introduced a source J (n) to these operators, namely, we write ∆L = J(n)µ1···µnO

(n)µ1···µnf,g , with

J(n)µ1···µn = Λµ1 · · ·Λµn , and Λ2 = 0. As can be easily seen, this source automatically projects

the operator O(n)f,g to its symmetric and traceless part.

(a) Firstly, we consider anfg, which can be got by evaluating the following two diagrams.

p

k

With the Feynman rules listed above, the first diagram reads,

(ig)2

∫d4k

(2π)4(−1) tr

[tbγν

i

/k/Λ

i

/ktaγµ

i

/k − /p

](Λ · k)n−1

=− ig2 tr [tatb]

∫d4k′

(2π)4

∫ 1

0

dx2(1− x)

(k′2 −∆)3(Λ · k)n−1 tr [γν/k/Λ/kγµ(/k − /p)]. (18.49)

We need to extract terms of proportional to gµν(Λ · p)n and of logarithmical divergence.

This needs some manipulations on the numerator of the integrand. We firstly evaluate

the gamma trace, keep terms containing at least two powers of k, and shift the variable

kµ → k′µ = kµ − xpµ. Then we pick up terms containing two k′, which contributes to

logarithmical divergence. At last we symmetrize the indices according to k′µk′ν → k′2gµν/4.

The detailed steps are given as follows.

(Λ · k)n−1 tr [γν/k/Λ/kγµ(/k − /p)]

⇒[16(Λ · k)nkµkν

]−[4(Λ · k)n(k − 2p) · kgµν

]−[4(Λ · k′)n−1(Λ · p)k2gµν

]⇒[16xn(Λ · p)nk′µk′ν

]−[4nxn(k′ · p)(Λ · k′)(Λ · p)n−1gµν

+ 4n(x− 2)xn−1(k′ · p)(Λ · k′)(Λ · p)n−1gµν + 4xn(Λ · p)nk′2gµν]

−[8(n− 1)xn−1(k′ · p)(Λ · k′)(Λ · p)n−1gµν + 4xn−1(Λ · p)nk′2gµν

]⇒[4xn](Λ · p)nk′2gµν −

[nxn + n(x− 2)xn−1 + 4xn

](Λ · p)nk′2gµν


−[2(n− 1)xn−1 + 4xn−1

](Λ · p)nk′2gµν

=− (2nxn + 2xn−1)(Λ · p)nk′2gµν . (18.50)

Then it is straightforward to finish the loop integral,

ig2 tr [tatb](Λ · p)ngµν∫ 1

0

dx 2(1− x)(2nxn + 2xn−1)

∫d4k′

(2π)4

k′2

(k′2 −∆)3

=− g2

(4π)2

2(n2 + n+ 2)

n(n+ 1)(n+ 2)

Γ(2− d2

)

∆2−d/2 (Λ · p)nδabgµν . (18.51)

The second diagram contributes an identical term for n even. The two diagrams sum to

g2

(4π)2

2(n2 + n+ 2)

n(n+ 1)(n+ 2)

Γ(2− d2

)

∆2−d/2 ·(− 2(Λ · p)nδabgµν

). (18.52)

Therefore the corresponding counterterm reads

δfg = − g2

(4π)2

2(n2 + n+ 2)

n(n+ 1)(n+ 2)

Γ(2− d2

)

(M2)2−d/2 , (18.53)

and the anomalous dimension element reads

γnfg = −M ∂

∂Mδfg = − g2

(4π)2

4(n2 + n+ 2)

n(n+ 1)(n+ 2), (18.54)

and thus,

anfg =4(n2 + n+ 2)

n(n+ 1)(n+ 2). (18.55)

(b) Then we consider angf and angg. This time we need to evaluate the following four

diagrams.

The first diagram contributes to angf , which reads

− 2(ig)2

∫d4k

(2π)4taγν

i

/p− /ktaγµ

( i

k2

)2

×[gµν(Λ · k)n + k2ΛµΛν(Λ · k)n−2 − 2k(µΛν)(Λ · k)n−1

]=− 2ig2C2(N)

∫d4k′

(2π)4

∫ 1

0

dx2(1− x)

(k′2 −∆)3

[γµ(/p− /k)γµ(Λ · k)n

+ /Λ(/p− /k)/Λ(Λ · k)n−2k2 −(/Λ(/p− /k)/k + /k(/p− /k)/Λ

)(Λ · k)n−1

](18.56)


To find the pieces proportional to /Λ(Λ ·p)n−1 and of logarithmical divergence, we manipulate

on the expression in the square bracket, shifting the variable kµ → k′µ = kµ−xpµ, extracting

terms with two factors of k′µ, symmetrizing the integrand with k′µk′ν → k′2gµν/4, and

throwing away terms proportional to Λ2(= 0). This gives[γµ(/p− /k)γµ(Λ · k)n

]+[/Λ(/p− /k)/Λ(Λ · k)n−2k2

]−[(/Λ(/p− /k)/k + /k(/p− /k)/Λ

)(Λ · k)n−1

]=[− 2(/p− /k)(Λ · k)n

]+[2/Λ(Λ · (p− k)

)(Λ · k)n−2k2

]−[2((

Λ · (p− k))/k +

((p− k) · k

)/Λ− (/p− /k)(Λ · k)

)(Λ · k)n−1

]⇒[2nxn−1/k

′(Λ · k′)(Λ · p)n−1

]+[− 2xn−1(k′ · p)(Λ · k′)(Λ · p)n−2

+ 2(n− 2)(1− x)xn−2(k′ · p)(Λ · k′)(Λ · p)n−2

+ (1− x)xn−2(Λ · p)n−1k′2]

+[2xn−1/k

′(Λ · k′)(Λ · p)n−1

− 2(n− 1)(1− x)xn−2(Λ · k′)(Λ · p)n−1 − 2/Λ(− xn−1k′2(Λ · p)n−1

− (n− 1)xn−1(k′ · p)(Λ · k′)(Λ · p)n−2

+ (n− 1)(1− x)xn−2(k′ · p)(Λ · k′)(Λ · p)n−2)− 2nxn−1/k

′(Λ · k′)(Λ · p)n−1

]⇒[ n

2xn−1

]/Λk′2(Λ · p)n−1 +

[− xn−1 + n(1− x)xn−2

]/Λk′2(Λ · p)n−1

+[− (n− 1)(1− x)xn−2 + 2xn−1

]/Λk′2(Λ · p)n−1

=[ n

2xn−1 + xn−2

]/Λk′2(Λ · p)n−1. (18.57)

Then we have

− 2ig2C2(N)/Λ(Λ · p)n−1

∫ 1

0

dx 2(1− x)( n

2xn−1 + xn−2

)∫ d4k′

(2π)4

k′2

(k′2 −∆)3

=g2C2(N)

(4π)2

2(n2 + n+ 2)

n(n2 − 1)

Γ(2− d2

)

∆2−d/2/Λ(Λ · p)n−1, (18.58)

which gives the counterterm coefficient,

δngf = − g2C2(N)

(4π)2

2(n2 + n+ 2)

n(n2 − 1)

Γ(2− d2

)

(M2)2−d/2 (18.59)

Then, in a similar way as in (a), we get

γngf = − g2C2(N)

(4π)2

4(n2 + n+ 2)

n(n2 − 1), (18.60)

and for N = 3, C2(N) = 4/3, we get

angf =16

3

(n2 + n+ 2)

n(n2 − 1). (18.61)


The second to fourth diagrams contribute to angg. Now we evaluate them in turn. The

second one reads,

− 2g2facef bdeδcd∫

d4k

(2π)4

( −i

k2

)2 −i

(p− k)2

×[gµρ(p+ k)λ + gρλ(p− 2k)µ + gλµ(k − 2p)ρ

]×[− gνσ(p+ k)λ + gσλ(2k − p)ν + gνλ(2p− k)σ

]×[gρσ(Λ · k)n + k2ΛρΛσ(Λ · k)n−2 − 2k(ρΛσ)(Λ · k)n−1

]⇒− 2ig2C2(G)δab

∫d4k′

(2π)4

∫ 1

0

dx2(1− x)

(k′2 −∆)3

[− 8(Λ · k)nkµkν −

((Λ · k)nk2

+ 2(Λ · k)n(k · p)− 8(Λ · k)n−1(Λ · p)(k · p) + 4(Λ · k)n−2(Λ · p)2k2)gµν]

⇒− 2ig2C2(G)gµνδab(Λ · p)n∫ 1

0

dx 2(1− x)[−(

3 +n

2

)xn − 1

2nxn−1 − 2xn−2

]×∫

d4k′

(2π)4

k′2

(k′2 −∆)3

=− g2C2(G)

(4π)2

(4

n+ 2− 6

n+ 1+

4

n− 4

n− 1

)Γ(2− d

2)

∆2−d/2 (−2)gµνδab(Λ · p)n. (18.62)

The third diagram reads (where an additional 1/2 is the symmetry factor),

− 1

2· 2ig2facdf bcd

∫d4k

(2π)4

−i

k2

−i

(p− k)2

×[gµρ(p+ k)σ + gρσ(p− 2k)µ + gσµ(k − 2p)ρ

]×

n∑j=1

[gνσΛρ(Λ · (p− k))j−1(Λ · p)n−j − gνρΛσ(Λ · p)j−1(Λ · k)n−j

]⇒− ig2C2(G)gµνδab(Λ · p)n

n∑j=1

∫ 1

0

dx[(1 + x)xn−j − (x− 2)(1− x)j−1

]×∫

d4k′

(2π)4

1

(k′2 −∆)2

=g2C2(G)

(4π)2gµνδab

Γ(2− 2d

)

∆2−d/2 (Λ · p)n

×n∑j=1

( 1

j+

1

j + 1+

1

n− j + 1+

1

n− j + 2

)⇒− g2C2(G)

(4π)2

[2

n∑j=2

1

j+

1

n+ 1+ 1

]Γ(2− 2

d)

∆2−d/2 (−2)gµν(Λ · p)n. (18.63)

The contribution from fourth diagram is identical to the one from the third diagram. Sum-

ming the last three diagram together, we get

g2C2(G)

(4π)2

(4

(n+ 1)(n+ 2)+

4

n(n+ 1)− 4

n∑j=2

1

j− 2

)Γ(2− d

2)

∆2−d/2 (−2)gµνδab(Λ · p)n. (18.64)


Thus the corresponding counterterm is

δg = − g2C2(G)

(4π)2

(4

(n+ 1)(n+ 2)+

4

n(n+ 1)− 4

n∑j=2

1

j− 2

)Γ(2− d

2)

(M2)2−d/2 . (18.65)

As a result,

γngg = M∂

∂M(−δg + δ3)

= − 2g2

(4π)2

[(4

(n+ 1)(n+ 2)+

4

n(n+ 1)− 4

n∑j=2

1

j− 1

3

)C2(G)− 4

3nfC(N)

], (18.66)

therefore, for N = 3, C2(N) = 4/3 and C(N) = 1/2, we have,

angg = 6

(4

(n+ 1)(n+ 2)+

4

n(n+ 1)− 4

n∑j=2

1

j− 1

3− 2

9nf

). (18.67)

18.4 Deep inelastic scattering from a photon

(a) The A-P equation for parton distributions in the photon can be easily written down by

using the QED splitting functions listed in (17.121) of Peskin & Schroeder. Taking account

of quarks’ electric charge properly, we have,

d

d logQfq(x,Q) =

3Q2qα

π

∫ 1

x

dz

z

Pe←e(z)fq

( xz,Q)

+ Pe←γ(z)fγ

( xz,Q)

, (18.68)

d

d logQfq(x,Q) =

3Q2qα

π

∫ 1

x

dz

z

Pe←e(z)fq

( xz,Q)

+ Pe←γ(z)fγ

( xz,Q)

, (18.69)

d

d logQfγ(x,Q) =

∑q

3Q2qα

π

∫ 1

x

dz

z

Pγ←e(z)

[fq

( xz,Q)

+ fq

( xz,Q)]

+ Pγ←γ(z)fγ

( xz,Q)

, (18.70)

where the splitting functions are

Pe←e(z) =1 + z2

(1− z)+

+3

2δ(1− z), (18.71)

Pγ←e(z) =1 + (1− z)2

z, (18.72)

Pe←γ(z) = z2 + (1− z)2, (18.73)

Pγ←γ(z) =− 2

3δ(1− z). (18.74)

We take q = u, d, c, s, and Qu,c = +2/3, Qd,s = −1/3. The factor 3 in the A-P equations

above takes account of 3 colors. Since no more leptons appear in final states other than

original e+e−, they are not included in the photon structure. With the initial condition

18.4. Deep inelastic scattering from a photon 157

fγ(x,Q0) = δ(1 − x) and fq,q(x,Q0) = 0 where Q0 = 0.5GeV, these distribution functions

can be solved from the equations above to the first order in α, to be

fq(x,Q) = fq(x,Q) =3Q2

qα

2πlog

Q2

Q20

[x2 + (1− x)2

], (18.75)

fγ(x,Q) =

(1−

∑q

Q2qα

πlog

Q2

Q20

)δ(1− x). (18.76)

(b) The formulation of deep inelastic scattering from a photon is similar to the one for the

proton, as described in Peskin & Schroeder. The process can be formulated as a two-photon

scattering, with one photon being hard and the other one play the role of proton, which

has the internal structure as shown in (a). Therefore, we can write down the corresponding

current product as

W µν = i

∫d4x eiq·x〈γ|TJµ(x)Jν(0)|γ〉, (18.77)

which can be again expanded in terms of scalar form factors,

W µν =(− gµν +

qµqν

q2

)W1 +

(P µ − qµ P · q

q2

)(P ν − qν P · q

q2

)W2. (18.78)

After operator product expansion, the form factor W2 can be expressed as

W2 = 3∑q

Q2q

∑n

8

Q2

(2q · P )n−2

(Q2)n−2Anq (Q2), (18.79)

and Anq (Q2) is a scale-dependent quantity, whose scaling behavior is dictated by the anoma-

lous dimension matrix γ. This matrix can be evaluated again from the diagrams in Fig. 18.13

in Peskin & Schroeder and in figures of last problem. The only difference is that we should

replace the gluon field with photon field. Therefore it is straightforward to see that anγγ = 0.

For anqq and anγq, we should take away the group factor C2(N) = 4/3, while for anqγ, we should

take away the factor tr (tatb) = δab/2. In addition, we should also include the factor Q2q

corresponding to electric charge of each quark. Then we have,

anqq =− 2Q2f

[1 + 4

n∑j=2

1

j− 2

n(n+ 1)

], (18.80)

anqγ =8Q2

f (n2 + n+ 2)

n(n+ 1)(n+ 2), (18.81)

anγq =4Q2

f (n2 + n+ 2)

n(n2 − 1), (18.82)

anγγ = 0. (18.83)

(c) The n = 2 moment photon structure function can be worked out through the moment

sum rules (18.154) in Peskin & Schroeder, where the matrix elements Anq in our case is

a scale-dependent quantity. This dependence can be found by evaluating the anomalous


dimension matrix of operator O(2)q as is done below (18.185) of Peskin & Schroeder, but

with different entries, given by

γ = − α

4π

a2uu 0 3× 2a2

uγ

0 a2dd 3× 2a2

uγ

a2γu a2

γd a2γγ

= − α

4π

− 6427

0 3227

0 − 1627

827

6427

1627

0

. (18.84)

(d) As can be inferred from (a), the photon structure function fγ(x,Q) is originally peaked

at x = 1 for Q = Q0, and the peak shifts toward smaller x and the peak goes lower and

broader, as Q goes large from Q0.

Chapter 19

Perturbation Theory Anomalies

19.1 Fermion number nonconservation in parallel E

and B fields

(a) In this problem we investigate the effect of chiral anomaly on the (non)conservation

of fermion number with definite chirality. Let us begin with the Adler-Bell-Jackiw anomaly

equation,

∂µjµ5 = − e2

16π2εµνρσFµνFρσ. (19.1)

Integrating the left hand side over the whole spacetime, we get the difference between the

numbers of right-handed fermions NR and of left-handed fermions NL, namely,∫d4x ∂µj

µ5 =

∫d4x ∂µ(jµR − j

µL) =

∫d3x (j0

R − j0L)∣∣∣t2t1

= ∆NR −∆NL, (19.2)

where we assume that the integral region for time is [t1, t2] and that ∂iji integrates to zero

with suitable boundary conditions (i.e. vanishing at spatial infinity or periodic boundary

condition). On the other hand,

εµνρσFµνFρσ = 4ε0ijkF0iFjk = −8F0i

(12εijkFjk

)= −8E ·B. (19.3)

Therefore, the ABJ anomaly equation gives,

∆NR −∆NL =e2

2π2

∫d4xE ·B. (19.4)

(b) The Hamiltonian for massless charged fermions with background electromagnetic field

is given by

H =

∫d3x

(πD0ψ − L

)= −

∫d3x iψγiDiψ, (19.5)

where π = iψ† is the canonical conjugate momentum of ψ, L = iψ /Dψ is the Lagrangian for

the fermion, and Dµ = ∂µ+ieAµ is the covariant derivative. Now we expand the Hamiltonian

159

160 Chapter 19. Perturbation Theory Anomalies

in the chiral basis,

H =−∫

d3x(ψ†L ψ†R

)(−iσ ·D 0

0 iσ ·D

)(ψLψR

)=

∫d3x

[ψ†L(iσ ·D)ψL − ψ†R(iσ ·D)ψR

]. (19.6)

(c) Now we focus on the eigenvalue problem of the right-handed fermion ψR, namely the

equation −iσ ·DψR = EψR. To be definite, we set the background electromagnetic potential

to be Aµ = (0, 0, Bx1, A) with B and A two constants. To seek the eigenfunction of the

form ψR =(φ1(x1), φ2(x1)

)Tei(k2x2+k3x3), we substitute it into the equation above and get

φ′1 = (k2 − eBx1)φ1 + i(E + k3 − eA)φ2, (19.7a)

φ′2 = i(E − k3 + eA)φ1 − (k2 − eBx1)φ2. (19.7b)

Eliminating φ2 from these two equations, we get a single differential equation in the form of

the harmonic oscillator,

φ′′1 −[e2B2

(x1 − k2

eB

)2

− E2 + (k3 − eA)2 − eB]φ1 = 0. (19.8)

(d) Now we specify the spatial boundary condition to be the box of length L in each side

and periodic boundary condition. Then the condition ψR(x1, x2, x3) = ψR(x1, x2 + L, x3) =

ψR(x1, x2, x3 + L) implies that k2 and k3 are quantized according to ki = 2πni/L (i = 2, 3).

On the other hand, k2 also has an upper bound since (19.8) shows that the center of the

oscillator would be out of the box if k2 is too large. This condition implies that k2/eB < L,

which further gives the maximum value of n2 to be (n2)max = eBL2/2π. Note also that

the energy eigenvalue does not depend on k2, thus each energy level consists of eBL2/2π

degenerate states. Furthermore, we can also write down explicitly the energy eigenvalue

associated with the state labeled by (n1, n3):

E = ±[( 2πn3

L− eA

)2

− (n1 + 32

)eB

]1/2

. (19.9)

(e) Now we consider the case with n1 = 0 for simplicity. Then the spectrum reads E =

2πn3/L− eA. Suppose that the background potential changes by ∆A = 2π/eL. Then it is

easy to see that all state with energy marked by n3 will turn to states with energy marked by

n3−1. Note that each energy eigenvalue is eBL2/2π-degenerate, thus the net change of right-

handed fermion number is −eBL2/2π. Similar analysis shows that the left-handed fermion

number get changed by eBL2/2π. Therefore the total change is ∆NR −∆NL = −eBL2/π.

19.2. Weak decay of the pion 161

19.2 Weak decay of the pion

(a) In this problem we study the decay of charged pion. So let us work out the amplitude

for π+ → `ν, with the effective four-fermion interaction

∆L =4GF√

2(¯Lγ

µνL)(uLγµdL) + h.c. (19.10)

and the relation

〈0|jµ5a(x)|πb(p)〉 = −ipµfπδabe−ip·x (19.11)

as inputs. Firstly we recall that

jµa = QLγµτaQL + QRγ

µτaQR, (19.12a)

jµ5a =− QγLµτ

aQL + QRγµτaQR, (19.12b)

where QL = (uL, dL)T and QR = (uR, dR)T . Thus,

1

2(jµ1 + ijµ2 − jµ51 − ijµ52) = QLγ

µ(τ 1 + iτ 2)QL = uLγµdL. (19.13)

Then we find the decay amplitude M(π+(p)→ `+(k)ν(q)

)to be

iM =4iGF√

2u(q)γµ

( 1− γ5

2

)v(k) · 1√

2fπipµ = −GFfπu(q)/p(1− γ5)v(k). (19.14)

(b) Now let us calculate the decay rate of the charged pion. We note that the amplitude

above can be further simplified to

iM = −GFfπu(q)(/q + /k)(1− γ5)v(k) = −GFfπmù(q)(1 + γ5)v(k). (19.15)

Therefore we have∑|M|2 = G2

Ff2πm

2` tr(/q(1 + γ5)(/k −m`)(1− γ5)

)= 8G2

Ff2πm

2`q · k, (19.16)

where the summation goes over all final spins. We choose the momenta to be

p = (mπ, 0, 0, 0), k = (Ek, 0, 0, k), q = (Eq, 0, 0,−k). (19.17)

Then the kinematics can be easily worked out to be

Ek =m2π +m2

`

2m2π

, Eq = k =m2π −m2

`

2mπ

(19.18)

The decay rate then follows straightforwardly,

Γ =1

2mπ

∫dΩ

16π2

k2

EkEq

( k

Ek+

k

Eq

)−1

· 8G2Ff

2πm

2`(q · k)

=G2Ff

2π

4πmπ

( m`

mπ

)2

(m2π −m2

`)2, (19.19)


and we have the ratio between two decay channels,

Γ(π+ → e+νe)

Γ(π+ → µ+νµ)=

m2e(m

2π −m2

e)2

m2µ(m2

π −m2µ)2' 10−4. (19.20)

Thus to determine the pion decay constant fπ, we can consider the channel µ+νµ only as

a good approximation. With the lifetime of charged pion τπ = 2.6× 10−8sec as well as mπ

and mµ, we find that

fπ =

√4πmπ

G2F τπ

( mπ

mµ

)(m2

π −m2µ)−1 ' 90.6MeV. (19.21)

19.3 Computation of anomaly coefficients

(a) By definition, Aabc = tr [ta, tb, tc]. Then for the product representation r1 × r2, we

have

Aabc(r1 × r2) = tr r1×r2

[ta ⊗ 1 + 1⊗ ta,

tb ⊗ 1 + 1⊗ tb, tc ⊗ 1 + 1⊗ tc

]= tr r1×r2

[ta ⊗ 1 + 1⊗ ta, tb, tc ⊗ 1 + tb ⊗ tc + tc ⊗ tb + 1⊗ tb, tc

]= tr r1×r2

([tatb, tc]⊗ 1 + [ta, tb]⊗ tc + [ta, tc]⊗ tb

+ tb ⊗ [ta, tc] + tb ⊗ [ta, tc] + 1⊗ [ta, tb, tc])

= tr r1 [ta, tb, tc] tr r2(1) + tr r2 [t

a, tb, tc] tr r1(1)

= Aabc(r1)d(r2) +Aabc(r2)d(r1). (19.22)

On the other hand, as we decompose the representation r1 × r2 into a direct product of

irreducible representations∑i

ri, we have

Aabc(∑

i

ri

)= tr Σr

[∑i

tai ,∑

j

tbj,∑k

tck

]= tr Σr

(∑i

∑j

∑k

[tai , tbj, tck])

=∑i

tr ri [tai , tbi , tci] =

∑i

Aabc(ri) (19.23)

Note that Aabc(r) = 12A(r)dabc where dabc is the unique symmetric gauge invariant. Then

equating the two expressions above, we get

d(r2)A(r1) + d(r1)A(r2) =∑i

A(ri). (19.24)

(b) In this part we show that the representation (3 × 3)a of SU(3) is equivalent to 3.

Let ψi be the base vectors of 3 representation. Then, a set of base vectors of (3 × 3)acan be chosen to be εijkψjψk. From the transformation rule ψi → Uijψj, we know that

the (3 × 3)a base vectors transform according to εijkψjψk → εimnUmjUnkψjψk. Now, it

19.3. Computation of anomaly coefficients 163

is easy to show that ε`mnUìUmjUnk is totally antisymmetric, and thus is proportional to

εijk. Let us write ε`mnUìUmjUnk = Cεijk, then taking U = I shows that C = 1. Now

we multiply both sides of this equality by (U †)ip. Since U is unitary, (U †)ip = (U−1)ip, so

we get εpmnUmjUnk = εijk(U†)ip. That is, the base vector εijkUjUk transforms according to

εijkUjUk → (U †)ìε`jkψjψk = (U∗)i`ε`jkψjψk, which is exactly the transformation rule of 3.

Now from A(3) = 1, it follows that A(3) = −1. Therefore A((3 × 3)a) = −1, and by

using the equation derived in (a), we have A((3× 3)s) = 6− (−1) = 7.

(c) Now we compute the anomaly coefficients for a and s representations of the SU(N)

group. As indicated in Peskin & Schroeder, it is enough to consider an SU(3) subgroup

of SU(N). Then the fundamental representation N is decomposed into a direct sum of

irreducible representations when restricted to SU(3), that is, N = 3 + (N − 3)1. This

decomposition is easily justified by considering the upper-left 3 × 3 block of a matrix in

fundamental representation of SU(N). When this block is treated as a transformation of

SU(3), the first three components of the vector on which the matrix acts form a fundamental

representation vector of SU(3), while the other (N − 3) components of the column vector

are obviously invariant. With this known, we have,

N×N =(3 + (N − 3)1

)×(3 + (N − 3)1

)= 3× 3 + 2(N − 3)3 + (N − 3)21. (19.25)

On the other hand, we know that N = s+a while s and a are irreducible. Then we have, by

(a), 2N ·A(N) = A(s)+A(a). But we already know that A(N) = 1. Thus A(s)+A(a) = 2N .

Now, to compute A(a), we make use of the SU(3) restriction,

(N×N)a = (3× 3)a + (N − 3)3 + 12

(N − 3)(N − 4)1. (19.26)

Then,

A(a) = A((3× 3)a

)+ (N − 3)A(3) = A(3) + (N − 3)A(3) = N − 4, (19.27)

and A(s) = 2N − A(a) = N + 4.

Now consider totally antisymmetric rank-j tensor representation. Again we decompose

the fundamental representation as N = 3+(n−3)1. Then the rank-j totally antisymmetric

tensor can be decomposed as

(Nj)a =(N − j) · · · (N − j + 1)

(j − 3)!(3× 3× 3)a +

(N − 3) · · · (N − j)(j − 2)!

(3× 3)a

+(N − 3) · · · (N − j − 1)

(j − 1)!3 + 1’s. (19.28)

Therefore,

A(a) =− (N − 3) · · · (N − j)(j − 2)!

+(N − 3) · · · (N − j − 1)

(j − 1)!

=(N − 3) · · · (N − j)(N − 2j)

(j − 1)!. (19.29)


19.4 Large fermion mass limits

In this problem we study the chiral anomaly and the trace anomaly in triangle diagrams

with Pauli-Villars regularization.

(a) and (c) Firstly we evaluate the expectation value of the divergence of the chiral current

jµ5 between the vacuum and the two-photon state, namely the matrix element 〈p, k|jµ5|0〉.This matrix element receives contributions at 1-loop level from the following two diagrams:

+

In momentum space, the divergence of the first diagram reads

iqµMµνλ1 = (−1)(−ie)2

∫d4`

(2π)4

tr

[/qγ

5 i

/− /kγλ

i

/γν

i

/+ /p

]− tr

[/qγ

5 i

/− /k −Mγλ

i

/−Mγν

i

/+ /p−M

](19.30)

The integral is finite, thus we are allowed to shift the integral variable. For the first trace

and the second trace above, we rewrite the /qγ5 factors, respectively, as follows,

/qγµ = (/+ /p− /+ /k)γµ = (/+ /p)γ

5 + γ5(/− /k),

/qγµ = (/+ /p−M − /+ /k +M)γµ = (/+ /p−M)γ5 + γ5(/− /k −M) + 2Mγ5.

Then, the loop integral becomes

iqµMµνλ1 = e2

∫d4`

(2π)4

tr

[γ5 1

/− /kγλ

1

/γν + γ5γλ

1

/γν

1

/+ /p

]− tr

[γ5 1

/− /k −Mγλ

1

/−Mγν + γ5γλ

1

/−Mγν

1

/+ /p−M

]+ 2M tr

[γ5 1

/− /k −Mγλ

1

/−Mγν

1

/+ /p−M

](19.31)

In the expression above, the first and the second lines are canceled by the corresponding

terms from the second diagram with (k, λ↔ p, ν), while the third line is doubled. Therefore

the sum of two diagrams gives

iqµMµνλ = 4e2M

∫d4`

(2π)4tr

[γ5 1

/− /k −Mγλ

1

/+Mγν

1

/+ /p−M

]= 4e2M

∫d4`

(2π)4

∫ 1

0

dx

∫ 1−x

0

dy2N1[

(`− xk + yp)2 −∆]3

=− 4ie2MN1

(4π)2

∫ 1

0

dx

∫ 1−x

0

dy1

∆(19.32)

19.4. Large fermion mass limits 165

with

N1 = tr[γ5(/− /k +M)γλ(/−M)γν(/+ /p+M)

]= −4iMεαβλνkαpβ,

∆ = M2 − x(1− x)k2 − y(1− y)p2 − 2xyk · p.

Then the integral can be carried out directly in the M2 →∞ limit, to be

iqµMµνλ =− e2

2π2εαβλνkαpβ, (19.33)

as expected.

(b) and (d) For scale anomaly, the diagrams are the same. Now the relevant matrix

element is given by 〈p, k|Mψψ|0〉. Then the first diagram reads

iMµνλ1 ε∗ν(p)ε

∗λ(k) = ie2M

∫d4`

(2π)4

[1

/− /k/ε∗(k)

1

/k/ε∗(p)

1

/+ /p

]− tr

[1

/− /k −M/ε∗(k)

1

/−M/ε∗(p)

1

/+ /p−M

](19.34)

The first trace vanishes upon regularization, then,


∗λ(k) =− ie2M

∫d4`′

(2π)4

∫ 1

0

dx

∫ 1−x

0

dy2N2

(`′2 −∆)3, (19.35)

where `′ = `− xk + yp, ∆ = M2 − 2xyk · p, and the trace in the numerator is

N2 = tr[(/− k +M)/ε∗(k)(/+M)/ε∗(p)(/+ /p+M)

]= 4M

[M2ε∗(k) · ε∗(p) +

(ε∗(k) · p

)(ε∗(p) · k

)−(ε∗(k) · ε∗(p)

)(k · p

)+ 4(ε∗(k) · `

)(ε∗(p) · `

)−(ε∗(k) · ε∗(p)

)`2]

= 4M[M2ε∗(k) · ε∗(p) + (1− 4xy)

(ε∗(k) · p

)(ε∗(p) · k

)− (1− 2xy)

(ε∗(k) · ε∗(p)

)(k · p) +

(4d− 1)(ε∗(k) · ε∗(p)

)`′2],

where we used the transverse condition k · ε∗(k) = p · ε∗(p) = 0, and in the last equality,

the substitution `′µ`′ν → 1dgµν`′2. We also dropped all terms linear in `′ in the last equality.

The integral is then divergent, and we regularize it by dimensional regularization. Then

after carrying out the loop integral, we get


∗λ(k) =

e2

4π2

[(ε∗(k) · ε∗(p)

)(k · p)−

(ε∗(k) · p

)(ε∗(p) · k

)]×∫ 1

0

dx

∫ 1−x

0

dy(1− 4xy)M2

M2 − 2xyk · p(19.36)

Then, taking M2 →∞ limit, we find


∗λ(k) =

e2

12π2

[(ε∗(k) · ε∗(p)

)(k · p)−

(ε∗(k) · p

)(ε∗(p) · k

)](19.37)

The second diagram is obtained, again, by the exchange (k, λ ↔ p, ν), which gives the

identical result. Therefore we finally get

iMµνλε∗ν(p)ε∗λ(k) =

e2

6π2

[(ε∗(k) · ε∗(p)

)(k · p)−

(ε∗(k) · p

)(ε∗(p) · k

)]. (19.38)


Chapter 20

Gauge Theories with Spontaneous

Symmetry Breaking

20.1 Spontaneous breaking of SU(5)

We consider two patterns of spontaneous breaking of SU(5) gauge symmetry, with an

adjoint-representation scalar field Φ picking up vacuum expectation values

〈Φ〉 = Adiag(1, 1, 1, 1,−4), 〈Φ〉 = Bdiag(2, 2, 2,−3,−3), (20.1)

respectively. The kinetic term of the scalar field in the Lagrangian is

Lkin. = tr(

(DµΦ)†(DµΦ))

= tr((∂µΦ + g[Aµ,Φ]

)†(∂µΦ + g[Aµ,Φ]

)). (20.2)

Then the mass term of gauge bosons after symmetry breaking is given by

∆L = g2 tr(

[Aµ,Φ]†[Aµ,Φ])

= −g2AaµAµb tr

([T a, 〈Φ〉][T b, 〈Φ〉]

). (20.3)

To analyze the gauge bosons’ spectrum, we note that there are 24 independent generators

for SU(5) group, each of which can be represented as a 5 × 5 traceless hermitian matrix.

Then, for the first choice of 〈Φ〉 = diag(1, 1, 1, 1,−4), we see that for the generators of the

form

T =

(T (4)

0

)and T = 1

2√

10diag(1, 1, 1, 1,−4),

where T (4) is a 4× 4 matrix being any generator of SU(4) group, the commutators vanish.

That is, a subgroup SU(4)× U(1) remains unbroken in this case. Then, for the rest of the

generators, namely

1

2

0 0 0 0 1

0 0 0 0 0

0 0 0 0 0

0 0 0 0 0

1 0 0 0 0

,1

2

0 0 0 0 i

0 0 0 0 0

0 0 0 0 0

0 0 0 0 0

−i 0 0 0 0

,1

2

0 0 0 0 0

0 0 0 0 1

0 0 0 0 0

0 0 0 0 0

0 1 0 0 0

,

167

168 Chapter 20. Gauge Theories with Spontaneous Symmetry Breaking

etc, it is easy to calculate the commutators to get the trace equal to −25A2/2. Thus the

corresponding components of gauge bosons acquire mass MA = 5gA. In the same way, we

can also analyze the case of 〈Φ〉 = diag(2, 2, 2,−3,−3). This time the unbroken subgroup is

SU(3) × SU(2) × U(1), and the remaining 12 components of gauge bosons acquire a mass

equal to MA = 5gB, as can be found by evaluating the corresponding commutators.

20.2 Decay modes of the W and Z bosons

(a) The relevant interaction term in the Lagrangian reads

∆L = 1√2gW+

µ

(∑i

νiLγµeiL +

∑j,c

ucjLγµdcjL

), (20.4)

where the sum on i goes over all three generations of leptons, the sum on j goes over the

first two generations of quarks, since mt > mW , and the sum on c is due to 3 colors.

Now consider the decay of W+ boson. The amplitude of the decay into a pair of fermions

is

iM =ig√

2εµ(k)u(p1)γµ

( 1− γ5

2

)v(p2), (20.5)

where εµ is the polarization vector for W+µ , and the labels for momenta are shown in Fig.

20.1. Thus the squared amplitude with initial polarizations averaged is

k

p2p1

W+

f f

k

p2p1

Z0

f f

Figure 20.1: The decay of W+ and Z0 into fermion-antifermion pairs. All initial momenta go

inward and all final momenta go outward.

1

3

∑spin

|iM|2 =g2

6

(− gµν +

kµkνm2W

)tr

[/p2γµ( 1− γ5

2

)/p1γν( 1− γ5

2

)]=

g2

3

(p1 · p2 + 2

k · p1k · p2

m2W

). (20.6)

The momenta in the center-of-mass frame can be taken to be

k = (mW , 0, 0, 0), p1 = (p, 0, 0, p), p2 = (p, 0, 0,−p), (20.7)

and energy conservation requires that p = mW/2. Thus we get

1

3

∑spin

|iM|2 =1

3g2m2

W , (20.8)

20.2. Decay modes of the W and Z bosons 169

and the decay rate∫dΓ =

1

2mW

∫d3p1d3p2

(2π)62E12E2

(1

3g2m2

W

)(2π)4δ(4)(k − p1 − p2) =

αmW

12 sin2 θw, (20.9)

where we have used g = e/ sin θw and α = e2/4π. For each quark final state we multiply

the result by a QCD correction factor(1 + αs

π

). Then, taking account of 3 generations of

leptons and 2 generations of quarks with 3 colors, we get the partial decay rate of W+ into

fermions,

Γ(W+ → e+i νi) =

αmW

12 sin2 θw' 0.23GeV; (20.10)

Γ(W+ → uj dj) =αmW

4 sin2 θw

(1 +

αsπ

)' 0.70GeV; (20.11)

Γ(W+ → fermions) =αmW

12 sin2 θw

(9 + 6

αsπ

)' 2.08GeV. (20.12)

and also the branching ratios BR(W+ → e+i νi) = 0.11%, and BR(W+ → uj dj) = 0.34%.

Note that the fine structure constant at mW is α(mW ) ' 1/129.

(b) In the same way, we can also calculate the decay rate of Z →fermions. The relevant

term in the Lagrangian is

∆L =g

cos θwZµ∑i

fiγµ(I3i − sin2 θwQi

)fi, (20.13)

where the sum goes over all left- and right-handed fermions, including 3 generations of

leptons, and the first two generations of quarks with 3 colors, while I3i and Qi are associated

3-component of the weak isospin and the electric charge, respectively.

Then we can write down the amplitudes of the decay of Z0 into a pair of fermions ff

with specific I3 and Q, as illustrated in Fig. 20.3,

iM =ig

cos θwεµ(k)u(p1)γµ

[(I3 − sin2 θwQ

)( 1− γ5

2

)− sin2 θwQ

( 1 + γ5

2

)]v(p2)

=ig

cos θwεµ(k)u(p1)γµ

[I3( 1− γ5

2

)− sin2 θwQ

]v(p2), (20.14)

the squared matrix elements,

1

3

∑spin

|iM|2 =g2

3 cos2 θw

(− gµν +

kµkνm2Z

)× tr

[/p2γµ(

12I3(1− γ5)− sin2 θwQ

)/p1γν(

12I3(1− γ5)− sin2 θwQ

)]=

4g2

3 cos θ2w

[(12I3 − sin2 θwQ

)2+(

12I3)2](p1 · p2 +

2k · p1k · p2

m2Z

)=

4g2m2Z

3 cos θ2w


)2+(

12I3)2], (20.15)


and the partial decay rate,

Γ(Z0 → ff) =αmZ

3 sin2 θw cos2 θw


)2+(

12I3)2]. (20.16)

We should also multiply the result for quarks by the QCD factor(1 + αs

π

). Now we list

the numerical results of partial width and the branching ratios for various decay products

as follows.

ff Γ(ff)/GeV BR(ff)

νeνe, νµνµ, ντ ντ 0.17 6.7%

e−e+, µ−µ+, τ−τ+ 0.08 3.4%

uu, cc 0.30 11.9%

dd, ss, bb 0.39 15.4%

All fermions 2.51 100%

20.3 e+e− →hadrons with photon-Z0 interference

(a) It is easier to work with amplitudes between initial and final fermions with definite

chirality. In this case the relevant amplitude is given by∗

iM = (ie)2v(k2)γµu(k1)−i

q2u(p1)γµQfv(p2)

[−Qf +

(I3e + s2

w)(I3f − s2

wQf )

s2wc

2w

q2

q2 −m2Z

],

(20.17)

where I3e = −1/2 or 0 when the initial electron is left-handed or right-handed, so as I3

f to the

final fermion. The momenta is labeled as shown in Fig. 20.2. Then we can find associated

k1

p2

k2

q

p1

e− e+

f f

Figure 20.2: The process of e+e− → ff via the exchange of a photon/Z0 in s-channel. The

directions of ki’s and pi’s are inward and outward, respectively.

differential cross section to be

dσ

d cos θ(e+Re−L → fRfL) =

πα2

2s(1 + cos θ)2FLL(f), (20.18a)

dσ

d cos θ(e+Re−L → fLfR) =

πα2

2s(1− cos θ)2FLR(f), (20.18b)

dσ

d cos θ(e+Le−R → fRfL) =

πα2

2s(1− cos θ)2FRL(f), (20.18c)

∗In this problem we simplify the notation by sw ≡ sin θw and cw ≡ cos θw.

20.3. e+e− →hadrons with photon-Z0 interference 171

dσ

d cos θ(e+Le−R → fLfR) =

πα2

2s(1 + cos θ)2FRR(f), (20.18d)

in which α is the fine structure constant, s = q2 is the center-of-mass energy, and the F

factors are defined as follows:

FLL(f) =

∣∣∣∣Qf +( 1

2− s2

w)(I3f − s2

wQf )

s2wc

2w

s

s−m2Z + imZΓZ

∣∣∣∣2, (20.19)

FLR(f) =

∣∣∣∣Qf −( 1

2− s2

w)Qf

c2w

s

s−m2Z + imZΓZ

∣∣∣∣2, (20.20)

FRL(f) =

∣∣∣∣Qf −(I3f − s2

wQf )

c2w

s

s−m2Z + imZΓZ

∣∣∣∣2, (20.21)

FRR(f) =

∣∣∣∣Qf +s2wQf

c2w

s

s−m2Z + imZΓZ

∣∣∣∣2, (20.22)

where we have added the correction from resonance by using the Breit-Wigner formula.

Summing up the four expressions in (20.18), averaging the initial spins, and integrating over

the angle θ, we get finally the unpolarized cross section

σ(ff) =πα2

3s

[FLL(f) + FLR(f) + FRL(f) + FRR(f)

]. (20.23)

When the final state particle f is a quark, one should multiply the result by 3(1+ αs

π

)where

3 is the color factor, and(1 + αs

π

)is the 1-loop QCD correction.

For the final fermion being muon (I3f = −1/2, Qf = −1), up quark (I3

f = 1/2, Q = 2/3),

and down quark (I3f = −1/2, Qf = −1/3), we plot the corresponding cross section as a

function of center-of-mass energy ECM =√s in Fig. 20.3.

50 100 150 200

10

100

1000

104

ECMGeV

Σp

b

Figure 20.3: The cross section σ(e+e− → ff) as a function of center-of-mass energy ECM. The

black, blue, and red curves correspond to ff = µ−µ+, uu and dd, respectively.

(b) Now we calculate the forward-backward asymmetry AfFB, defined to be

AfFB =σF − σBσF + σB

=

( ∫ 1

0−∫ 0

−1

)d cos θ(dσ/d cos θ)( ∫ 1

0+∫ 0

−1

)d cos θ(dσ/d cos θ)

. (20.24)


Then from (20.18), we find

σF =πα2

24s

[7FLL(f) + FLR(f) + FRL(f) + 7FRR(f)

], (20.25)

σB =πα2

24s

[FLL(f) + 7FLR(f) + 7FRL(f) + FRR(f)

]. (20.26)

Thus

AfFB =3

4· FLL(f)− FLR(f)− FRL(f) + FRR(f)

FLL(f) + FLR(f) + FRL(f) + FRR(f). (20.27)

Again, we plot AfFB, as a function of ECM, for f = µ−, u, d, in Fig. 20.4.

MZ

40 60 80 100 120 140

-0.6

-0.4

-0.2

0.0

0.2

0.4

0.6

ECMGeV

AF

B

Figure 20.4: The forward-backward asymmetry AfFB as a function of center-of-mass energy ECM.

The black, blue, and red curves correspond to ff = µ−µ+, uu and dd, respectively.

(c) Recall the definition of F ’s, we find, on the Z0 resonance (s = mZ),

FLL(f) '[ ( 1

2− s2

w

)(I3f − s2

wQf

)s2wc

2w

mZ

ΓZ

]2

, FLR(f) '[ ( 1

2− s2

w

)Qf

c2w

mZ

ΓZ

]2

,

FRL(f) '[ (I3f − s2

wQf

)c2w

mZ

ΓZ

]2

, FRR(f) '[s2wQf

c2w

mZ

ΓZ

]2

,

therefore,

AfFB =3

4·[(

12− s2

w

)2 − s4w

][(I3f − s2

wQf

)2 − (s2wQf )

2][(

12− s2

w

)2+ s4

w

][(I3f − s2

wQf

)2+ (s2

wQf )2] =

3

4AeLRA

fLR. (20.28)

(d)

σpeak =πα2

3m2Z

· 1

s4wc

4w

· m2Z

Γ2Z

[(12− s2

w

)2+ s4

w

][(I3f − s2

wQf

)2+ (s2

wQf )2]

=12π

m2ZΓ2

Z

(αmZ

6s2wc

2w

[(12− s2

w

)2+ s4

w

])( αmZ

6s2wc

2w

[(I3f − s2

wQf

)2+ (s2

wQf )2])

=12π

m2Z

· Γ(Z0 → e+e−)Γ(Z0 → ff)

Γ2Z

. (20.29)

20.4. Neutral-current deep inelastic scattering 173

20.4 Neutral-current deep inelastic scattering

(a) In this problem we study the neutral-current deep inelastic scattering. The process is

mediated by Z0 boson. Assuming mZ is much larger than the energy scale of the scattering

process, we can write down the corresponding effective operators, from the neutral-current

Feynman rules in electroweak theory,

∆L =g2

4m2W

(νγµ)PLν

[uγµ

((1− 4

3s2w

)PL − 4

3s2wPR

)u

+ dγµ

((1− 2

3s2w)PL − 2

3s2wPR

)d]

+ h.c., (20.30)

where PL = (1−γ5)/2 and PR = (1+γ5)/2 are left- and right-handed projectors, respectively.

Compare the effective operator with the charged-operator in (17.31) of Peskin & Schroeder,

we can write down directly the differential cross section by modifying (17.35) in Peskin &

Schroeder properly, as

d2σ

dxdy(νp→ νX) =

G2F sx

4π

[(1− 4

3s2w

)2+ 16

9s4w(1− y2)

]fu(x)

+[(

1− 23s2w

)2+ 4

9s4w(1− y2)

]fd(x)

+[

169s4w +

(1− 4

3s2w

)2(1− y2)

]fu(x)

+[

49s4w +

(1− 2

3s2w

)2(1− y2)

]fd(x)

, (20.31)

d2σ

dxdy(νp→ νX) =

G2F sx

4π

[169s4w +

(1− 4

3s2w

)2(1− y2)

]fu(x)

+[

49s4w +

(1− 2

3s2w

)2(1− y2)

]fd(x)

+[(

1− 43s2w

)2+ 16

9s4w(1− y2)

]fu(x)

+[(

1− 23s2w

)2+ 4

9s4w(1− y2)

]fd(x)

. (20.32)

(b) For the neutrino scattering from a nucleus A with equal numbers of protons and

neutrons, we have fu = fd and fu = fd. Then the differential cross sections reads

d2σ

dxdy(νA→ νX) =

G2F sx

π

[12− s2

w + 59s4

4 + 59s4

4(1− y2)]fu(x)

+[

59s4w +

(12− s2

w + 59s4w

)(1− y2)

]fu(x)

, (20.33)

d2σ

dxdy(νp→ νX) =

G2F sx

π

[59s4w +

(12− s2

w + 59s4w

)(1− y2)

]fu(x)

+[

12− s2

w + 59s4

4 + 59s4

4(1− y2)]fu(x)

. (20.34)

Recall that for charged-current neutrino deep inelastic scattering, the differential cross sec-

tions are given by (17.35) in Peskin & Schroeder. Thus it is easy to find that

Rν =d2σ/dxdy(νA→ νX)

dσ/dxdy(νA→ µ−X)=

1

2− s2

w +5

9s4w

(1 +

fu(x)(1− y2) + fu(x)

fu(x) + fu(x)(1− y)2

), (20.35)


sin2Θw = 0.23

0.0 0.1 0.2 0.3 0.4 0.50.2

0.4

0.6

0.8

1.0

RΝ

RΝ

Figure 20.5: Weinberg’s nose with r = 0.4. See problem 21.4.

Ru =d2σ/dxdy(νA→ νX)

dσ/dxdy(νA→ µ+X)=

1

2− s2

w +5

9s4w

(1 +

fu(x) + fu(x)(1− y)2

fu(x)(1− y2) + fu(x)

), (20.36)

wherefu(x)(1− y2) + fu(x)

fu(x) + fu(x)(1− y)2= r. (20.37)

(c) The plot “Weinberg’s Nose” with r = 0.4 is shown in Figure 20.5.

20.5 A model with two Higgs fields

(a) The gauge boson mass matrix comes from the kinetic term of scalar fields,

(Dµφ1)†(Dµφ1) + (Dµφ2)†(Dµφ2),

with Dµφ1,2 =(∂µ− i

2gAaµσ

a− i2g′Bµ

)φ1,2. After φ1,2 acquire the vacuum expectation value

1√2

(0v1,2

), we observe that each of the kinetic terms gives rise to mass terms for gauge bosons

similar to the ones in the standard electroweak theory. Thus it is straightforward that the

masses of gauge bosons in this model is given by the replacement v2 → v21 + v2

2.

(b) The statement that the configuration 1√2

(0v1,2

)is a locally stable minimum, is equivalent

to that all particle excitations generated above this solution have positive squared mass m2.

Thus we investigate the mass spectrum of the theory with the vacuum chosen to be 1√2

(0v1,2

).

Firstly, we parameterize two scalar doublets as

φi =

(π+i ,

1√2(vi + hi + iπ0

i )

), (i = 1, 2) (20.38)

and substitute this parameterization into the potential,

V =− µ21φ†1φ1 − µ2

2φ†2φ2 + λ1(φ†1φ)2 + λ2(φ†2φ2)2

+ λ3(φ†1φ1)(φ†2φ2) + λ4(φ†1φ2)(φ†2φ1) + λ5

((φ†1φ2)2 + h.c.

). (20.39)

20.5. A model with two Higgs fields 175

Then the mass term of various scalar components can be extracted, as follows.

Lmass = (λ4 + 2λ5)v1v2

(π−1 π−2

)(v2/v1 −1

−1 v1/v2

)(φ+

1

φ+2

)

+ 2λ5v1v2

(π0

1 π02

)(v2/v1 −1

−1 v1/v2

)(φ0

1

φ02

)

− v1v2

(h1 h2

)( λ1(v1/v2) λ3 + λ4 + 2λ5

λ3 + λ4 + 2λ5 λ2(v2/v1)

)(h1

h2

). (20.40)

The eigenvalues of these matrices are easy to be found. For charged components, there

is a zero mode corresponding two broken directions in SU(2), and the mass of the other

charged scalar is given by m2c = −(λ4 + 2λ5)(v2

1 + v22). For pseudoscalar components, there

is also a zero mode corresponding to the rest one direction of broken SU(2), and the mass

of the other pseudoscalar is m2p = −4λ5(v2

1 + v22). Finally, for neutral scalars, the two mass

eigenvalues are given by the roots of following equation,

m2n − (λ1v

21 + λ2v

22)m2

n +[λ1λ2 − (λ3 + λ4 + 2λ5)2

]= 0. (20.41)

Therefore, to make m2c > 0, m2

p > 0 and m2n > 0, it is sufficient that

λ4 + 2λ5 < 0, λ5 < 0, λ1, λ2 > 0, λ1λ2 > (λ3 + λ4 + 2λ5)2. (20.42)

(c) From the mass terms in (b) we can diagonalize the charged scalar mass matrix with

the rotation matrix (π+

φ+

)=

(cos β sin β

− sin β cos β

)(π+

1

π+2

), (20.43)

where π+ is the Goldstone mode and φ+ is a physical charged scalar. Given that φ+ to get

the physical mass, it is easy to see that the rotation angle can be chosen to be tan β = v2/v1.

(d) Assuming that the Yukawa interactions between quarks and scalars take the following

form,

Lm = −(uL dL

)[λd

(π+

11√2v1

)dR + λu

(1√2v2

π−

)uR

]+ h.c., (20.44)

where we have suppressed flavor indices and neglected neutral scalar components. We

focus on charged component only. Then, with Peskin & Schroeder’s notation, we make the

replacement uL → UuuL, dL → UddL, uR → WuuR, and dR → WddR. Then, together with

λd = UdDdW†d and λu = UuDuW

†u where Dd and Du are diagonal matrix, we have

Lm =− 1√2

(v1dLDddR + v2uLDuuR

)− uVCKMDddRπ

+1 + dLV

†CKMDuuRπ

−2 + h.c. (20.45)


From the first line we see that the diagonal mass matrix for quarks are given by mu =

(v1/√

2)Du and md = (v2/√

2)Dd. We further define v =√v2

1 + v22 and note that π+

1 =

−φ+ sin β · · · , π+2 = φ+ cos β + · · · , then the Yukawa interactions between charged boson

and quarks can be written as

Lm ⇒−√

2

v1

(uLVCKMmddRπ

+1 + dLV

†CKMmuuRπ

−2

)+ h.c.

⇒√

2

v

(uLVCKMmddRφ

+ tan β + dLV†

CKMmuuRφ− cot β

)+ h.c.. (20.46)

Chapter 21

Quantization of Spontaneously

Broken Gauge Theories

21.1 Weak-interaction contributions to the muon g− 2

In this problem we study the weak-interaction corrections to the muon’s anomalous

magnetic moment (AMM). The relevant contributions come from the W -neutrino loop and

Z-muon loop, together with the diagrams with the gauge bosons replaced with the cor-

responding Goldstone bosons. Here we will evaluate the W -neutrino loop diagram with

Feynman-’t Hooft gauge and general Rξ gauge in part (a) and part (b) respectively, and

Z-muon diagram in part (c).

(a) Now we come to the W -neutrino loop diagram and the corresponding Goldstone boson

diagrams, shown in Fig. 21.1.

q

k

q + k µ−

µ+

(a)

q

µ−

µ+

(b)

q

µ−

µ+

(c)

q

µ−

µ+

(d)

Figure 21.1: The weak-interaction contributions to muon’s EM vertex. These four diagrams

contain neutrino internal lines in the loops.

The Fig. 21.1(a) with W -neutrino loop reads

δ(a)ν Γµ(q) =

(ig)2

2

∫d4k

(2π)4

[gρλ(2k + q)µ + gλµ(−2q − k)ρ + gρµ(q − k)λ

]× −igρσk2 −m2

W

−igλκ(q + k)2 −m2

W

u(p′)γσ( 1− γ5

2

) i

/p′ + /kγκ( 1− γ5

2

)u(p)

=ig2

2

∫d4k′

(2π)4

∫ 1

0

dx

∫ 1−x

0

dy2

(k′2 −∆)3u(p′)

[(2k + q)µγσ(/p

′ + /k)γσ

177

178 Chapter 21. Quantization of Spontaneously Broken Gauge Theories

+ (−/k − 2/q)(/p′ + /k)γµ + γµ(/p

′ + /k)(/q − /k)]( 1− γ5

2

)u(p), (21.1)

where

k′ = k + xq + yp′,

∆ = (1− y)m2W − x(1− x)q2 − y(1− y)p′2 + 2xyq · p′.

To extract the form factor F2(q2), recall that the total diagram can be written as a linear

combination of (p′+p)µ, qµ, γµ, and parity-violating terms containing γ5. Only the (p′+p)µ

terms contribute to F2(q2) through the Gordon identity. With this in mind, now we try to

simplify the expression in the square bracket in (21.1), during which we will drops terms

proportional to qµ or γµ freely, and totally ignore the γ5 terms.[(2k + q)µγσ(/p

′ + /k)γσ

]+[(−/k − 2/q)(/p

′ + /k)γµ]

+[γµ(/p

′ + /k)(/q − /k)]

=[− 2(2k′ + (1− 2x)q − 2yp′

)µ(/k′ − x/q + (1− y)/p

′)]+[(− /k′ − (2− x)/q + y/p

′)(/k′ − x/q + (1− y)/p′)γµ]

+[γµ(/k′ − x/q + (1− y)/p

′)(− /k′ + (1 + x)/q + y/p′)]

⇒[4y(1− y)mp′µ

]+[2(x+ 2y − 2)mpµ

]+[2(−1− x+ y)mp′µ

]⇒ − (1− y)(3− 2y)m(p′ + p)µ

⇒ 2(1− y)(3− 2y)m2 · iσµνqν2m

.

The steps of this calculation is basically in parallel with the one of Problem 7.2. Here we have

written the mass of muon as m instead of mµ to avoid confusions. Thus the contribution to

the muon’s AMM from Fig. 21.1(a) is

ig2

2

∫d4k

(2π)4

∫ 1

0

dx

∫ 1−x

0

dy2

(k′2 −∆)3· 1

2· 2(1− y)(3− 2y)m2

' 7

3· g2m2

64π2m2W

=7

3· GFm

2

8π2√

2, (21.2)

where we have used the approximation mW m, and set q2 = 0 in the second line. The

Fermi constant GF/√

2 = g2/8m2W .

Fig. 21.1(b) and 21.1(c) read

δ(b)δ Γµ(q) =

ig√2· −mW

2· −i√

2gm

mW

∫d4k

(2π)4gµρ

i

k2 −m2W

−igρσ(q + k)2 −m2

W

× u(p′)( 1− γ5

2

) i

/p′ + /kγσ( 1− γ5

2

)u(p). (21.3)

δ(c)δ Γµ(q) =

ig√2· −mW

2· −i√

2gm

mW

∫d4k

(2π)4gµρ

−igρσk2 −m2

W

i

(q + k)2 −m2W

21.1. Weak-interaction contributions to the muon g − 2 179

× u(p′)γσ( 1− γ5

2

) i

/p′ + /k

( 1 + γ5

2

)u(p). (21.4)

Through the calculation similar to that of Fig. 21.1(a), it is easy to show that these two

diagrams contribute the same to the AMM, which reads

1

2· GFm

2

8π2√

2. (21.5)

Finally, Fig. 21.1(d) reads

δ(d)δ Γµ(q) =

( −i√

2gm

mW

)2∫

ddk

(2π)d(2k − q)µ i

k2 −m2W

i

(q + k)2 −m2W

× u(p′)( 1− γ5

2

) i

/p′ + /k

( 1 + γ5

2

)u(p). (21.6)

But it is not difficult to see that the contribution to the muon’s AMM from this diagram

is proportional to (m/mW )4, which can be omitted in the limit mW m, compared with

the other three diagrams. Therefore we conclude that the AMM of the muon contributed

by W -neutrino and corresponding Goldstone boson’s 1-loop diagrams is

aµ(ν) =

[7

3+

1

2+

1

2+O

( m2

m2W

)]· GFm

2

8π2√

2' 10

3· GFm

2

8π2√

2. (21.7)

(c) Now we come to the second set of diagrams as shown in Fig. 21.2.

q

µ−

µ+

(a)

q

µ−

µ+

(b)

Figure 21.2: The weak-interaction contributions to muon’s EM vertex. These two diagrams

contains no neutrino internal lines.

Firstly the Fig. 21.2(a) reads

δ(a)Z Γµ(q) =

( ig

4cw

)2∫

ddk

(2π)d−igρσ

(p′ + k)2 −m2Z

× u(p′)γρ(4s2

w − 1− γ5) i

−/k −mγµ

i

−/q − /k −mγσ(4s2

w − 1− γ5)u(p)

⇒ −ig2

16c2w

∫ddk′

(2π)d

∫ 1

0

dx

∫ 1−x

0

dy2

(k′2 −∆)3u(p′)γρ

[(/k +m)γµ(/k + /q +m)

+ (4s2w − 1)2(/k −m)γµ(/k + /q −m)

]γρu(p), (21.8)


where we have omitted terms proportional to γ5, as indicated by “⇒” sign, and

k′ = k + xq + yp′,

∆ = (1− y)m2 + ym2Z − x(1− x)q2 − y(1− y)p′2 + 2xyq · p′.

We will again focus only on the terms proportional to (p′ + p)µ. Then the spinor part can

be reduced to

u(p′)γρ[(/k +m)γµ(/k + /q +m) + (3s2

w − c2w)2(/k −m)γµ(/k + /q −m)

]γρu(p)

⇒[2y(3 + y)− (4s2

w − 1)2 · 2y(1− y)]2m2 · u(p′)

iσµνqν2m

u(p).

Thus the AMM contribute by this diagram is

− ig2

16c2w

∫d4k′

(2π)4

∫ 1

0

dx

∫ 1−x

0

dy2 · 2m2[2y(3 + y)− (4s2

w − 1)2 · 2y(1− y)]

(k′2 −∆)3

=GFm

2

8π2√

2· 1

3

[(4s2

w − 1)2 − 5]. (21.9)

On the other hand, the Fig. 21.2(b) only contributes terms of order m4/m4W that can be

omitted, as can be seen from the coupling between the Goldstone boson and the muon.

Thus we conclude that the total contribution to aµ(Z) from the two diagrams in Fig. 21.2

at the leading order is given by (21.9).

21.2 Complete analysis of e+e− → W+W−

In this problem we calculate the amplitude for the process e+e− → W+W− at tree level

in standard electroweak theory. There are 3 diagrams contributing in total, as shown in

Figure 21.3.

k1

p2

k2

γp1

e− e+

W+ W−

Z0

e− e+

W+ W−

ν

e− e+

W+ W−

Figure 21.3: The process e−e+ → W+W− at tree level. All initial momenta go inward and all

final momenta go outward.

We will evaluate these diagrams for definite helicities for initial electrons as well as

definite polarizations for finalW bosons. The initial and final momenta can be parameterized

as

kµ1 = (E, 0, 0, E), pµ1 = (E, p sin θ, 0, p cos θ),

kµ2 = (E, 0, 0,−E), pµ2 = (E,−p sin θ, 0,−p cos θ), (21.10)

21.2. Complete analysis of e+e− →W+W− 181

with E2 = p2 +m2W , and electron mass ignored. For initial electron and positron, the spinors

with definite helicities can be chosen to be

uL(k1) =√

2E(0, 1, 0, 0)T , vL(k2) =√

2E(1, 0, 0, 0)T ,

uR(k1) =√

2E(0, 0, 1, 0)T , vR(k2) =√

2E(0, 0, 0, 1)T . (21.11)

For final W bosons, the polarization vectors are

ε∗−µ(p1) = 1√2(0,− cos θ,−i, sin θ), ε∗−µ(p2) = 1√

2(0, cos θ,−i,− sin θ),

ε∗+µ(p1) = 1√2(0,− cos θ, i, sin θ), ε∗+µ(p2) = 1√

2(0, cos θ, i,− sin θ),

ε∗Lµ(p1) = 1mW

(p,−E sin θ, 0,−E cos θ), ε∗Lµ(p2) = 1mW

(p, E sin θ, 0, E cos θ). (21.12)

It is easy to see that for initial electron-positron pair, only two helicity states e−Le+R

and e−Re+L contribute nonzero amplitudes. This is because the first two diagrams with s-

channel gauge bosons vanish for the other two possibilities e−Le+L and e−Re

+R due to angular

momentum conservation, while the third diagram vanishes since the weak coupling vanishes

for right-handed electron and left-handed positron. With this known, we can write down

the amplitudes for e−Le+R and e−Re

+L initial states, as follows. Generally the amplitude reads

iM(e−Le+R → W+W−)

=

[(−ie)

−i

(k1 + k2)2(ie) +

ie(− 12

+ s2w)

cwsw

−i

(k1 + k2)2 −m2Z

(igcw)

]× vL(k2)γλuL(k1)

[ηµν(p2 − p1)λ + ηνλ(−p1 − 2p2)µ + ηλµ(2p1 + p2)ν

]+( ig√

2

)2

vL(k2)γµi

/k1 − /p2

γνuL(k1)

ε∗µ(p1)ε∗ν(p2)

= ie2

[m2Z

s(s−m2Z)− 1

2s2w

1

s−M2Z

]vL(k2)

(ε∗(p1) · ε∗(p2)(/p2

− /p1)

− (p1 + 2p2) · ε∗(p1)/ε∗(p2) + (2p1 + p2) · ε∗(p2)/ε∗(p1))uL(k1)

− ie2

2s2w

1

u· vL(k2)/ε∗(p1)(/k1 − /p2

)/ε∗(p2)uL(k1), (21.13)

and,

iM(e−Re+L → W+W−)

=

[(−ie)

−i

(k1 + k2)2(ie) +

ies2w

cwsw

−i

(k1 + k2)2 −m2Z

(igcw)

]vR(k2)γλuR(k1)

×[ηµν(p2 − p1)λ + ηνλ(−p1 − 2p2)µ + ηλµ(2p1 + p2)ν

]ε∗µ(p1)ε∗ν(p2)

= ie2 m2Z

s(s−m2Z)vR(k2)

(ε∗(p1) · ε∗(p2)(/p2

− /p1)− (p1 + 2p2) · ε∗(p1)/ε∗(p2)

+ (2p1 + p2) · ε∗(p2)/ε∗(p1))uR(k1), (21.14)


In what follows we need the inner products among some of these vectors, as listed below.

p1 · p2 = E2 + p2 p1 · ε∗0(p2) = p2 · ε∗0(p1) =2Ep

mW

,

ε∗+(p1) · ε∗+(p2) = ε∗−(p1) · ε∗−(p2) = 1, ε∗0(p1) · ε∗0(p2) =E2 + p2

m2W

. (21.15)

We also need

vL(k2)/p1uL(k1) = −vL(k2)/p2

uL(k1) = 2Ep sin θ, (21.16)

vL(k2)/ε∗±(p1)uL(k1) = 2E( ∓1 + cos θ√

2

), (21.17)

vL(k2)/ε∗±(p2)uL(k1) = 2E( ∓1− cos θ√

2

), (21.18)

vL(k2)/ε∗0(p1)uL(k1) = −vL(k2)/ε∗0(p2)u(k1) =2E2 sin θ

mW

, (21.19)

vR(k2)/p1uR(k1) = −vR(k2)/p2

uR(k1) = −2Ep sin θ, (21.20)

vL(k2)/ε∗±(p1)uL(k1) = −2E( ∓1 + cos θ√

2

), (21.21)

vL(k2)/ε∗±(p2)uL(k1) = −2E( ∓1− cos θ√

2

), (21.22)

vL(k2)/ε∗0(p1)uL(k1) = −vL(k2)/ε∗0(p2)u(k1) = − 2E2 sin θ

mW

. (21.23)

We first consider e−Le+R → W+W−. In this case we take u(k1) = uL(k1) =

√2E(0, 1, 0, 0)T

and v(k2) = vL(k2) =√

2E(0, 0, 1, 0). Then each of the final W particle can have polar-

ization (+,−, 0), which gives 9 possible combinations for (W+,W−). Now we evaluate the

corresponding amplitudes in turn.

iM(e−Le+R → W+

(0)W−(0))

= ie2

[m2Z

s(s−m2Z)− 1

2s2w

1

s−M2Z

](− 4Ep(E2 + p2)

m2W

+16E3p

m2W

)sin θ

+ie2

2s2w

1

u· 2E(−3E2p+ p3 − 2E3 cos θ) sin θ

m2W

=− ie2 · s

4m2W

m2Z

s−m2Z

· β(3− β2)

− 1

2s2w

[( 2

1 + β2 + 2β cos θ− s

s−m2Z

)β(3− β2) +

4 cos θ

1 + β2 + 2β cos θ

]sin θ (21.24)

iM(e−Le+R → W+

(0)W−(±)) = iM(e−Le

+R → W+

(∓)W−(0))

= ie2

[m2Z

s(s−m2Z)− 1

2s2w

1

s−M2Z

]( 8E2p

mW

∓1 + cos θ√2

)− ie2

2s2w

1

u· 2E

mW

(E2(2 cos θ ∓ 1) + 2Ep± p2)±1 + cos θ√

2

21.3. Cross section for du→W−γ 183

= ie2

[m2Z

s−m2Z

β − 1

2s2W

(s

s−m2Z

β +±1− 2 cos θ − 2β ∓ β2

1 + β2 + 2β cos θ

)] √s

mW

±1 + cos θ√2

(21.25)

iM(e−Le+R → W+

(±)W−(±))

= ie2

[m2Z

s(s−m2Z)− 1

2s2w

1

s−M2Z

](− 4Ep sin θ

)+

ie2

2s2w

1

u· 2E(p+ E cos θ) sin θ

= ie2

[− m2

Z

(s−m2Z)β +

1

2s2w

(s

s−M2Z

β − 2(β + cos θ)

(1 + β2 + 2β cos θ)

)]sin θ (21.26)

iM(e−Le+R → W+

(±)W−(∓))

=− ie2

2s2w

1

u· 2E2(∓1 + cos θ) sin θ =

ie2

2s2w

2(±1− cos θ) sin θ

(1 + β2 + 2β cos θ). (21.27)

Though not manifest, these expressions have correct high energy behavior. To see this,

we note that β ' 1 − 2m2W/s when s m2

W . Then, for instance, the amplitude for two

longitudinal W final state becomes

iM(e−Le+R → W+

(0)W−(0)) = −ie2 · s

4m2W

m2Z

s−m2Z

· β(3− β2)

− 1

2s2w

[( 2

1 + β2 + 2β cos θ− s

s−m2Z

)β(3− β2) +

4 cos θ

1 + β2 + 2β cos θ

]sin θ

= − ie2

2s2W

(1 + 2 cos θ) sin θ

1 + cos θ+O(1/s). (21.28)

Then we can plot the azimuthal distribution of the corresponding differential cross section

at s = (1000GeV)2, as shown in Figure 21.4.

Next we consider the other case with e−Re+L initial state. Now there is no contribution

from u-channel neutrino exchange. The amplitudes for various polarizations of final W pairs

can be worked out to be

iM(e−Re+L → W+

(0)W−(0)) = ie2 s

s−m2Z

m2Z

4m2W

β(β2 − 3) sin θ, (21.29)

iM(e−Re+L → W+

(0)W−(±)) = iM(e−Re

+L → W+

(∓)W−(0))

= ie2 m2Z

s−m2Z

√s

mW

β±1− cos θ√

2(21.30)

iM(e−Re+L → W+

(±)W−(±)) = ie2 m2

Z

s−m2Z

β sin θ (21.31)

iM(e−Re+L → W+

(±)W−(∓)) = 0. (21.32)

21.3 Cross section for du→ W−γ

In this problem we compute the tree amplitude of du → W−γ at high energies so that

the quark masses can be ignored. In this case the left-handed and right-handed spinors


-1.0 -0.5 0.0 0.5 1.00.01

0.1

1

10

100

cos Θ

dΣd

co

sΘ

total

H0,0L

H+,-L

H-,+L

H+,0L+H0,-LH-,0L+H0,+L

Figure 21.4: The differential cross section of e−e+ → W+W− with definite helicity as a

function of azimuthal angle at s = (1000GeV)2.

k1

p2

k2

p1

d u

γ W−

k1

p2p1

k2d u

γ W−

k1

p1

k2

p2

d u

γ W−

Figure 21.5: The process du → W−γ at tree level. All initial momenta go inward and all final

momenta go outward.

decouple and only the amplitudes with dLuR initial state do not vanish. To calculate it, we

firstly work out the kinematics as follows.

k1 = (E, 0, 0, E), p1 = (p, p sin θ, 0, p cos θ),

k2 = (E, 0, 0,−E), p2 = (EW ,−p sin θ, 0,−p cos θ), (21.33)

where p = E−m2W/4E and EW = E +m2

W/4E. The initial spinors of definite helicities are

given by

uL(k1) =√

2E(0, 1, 0, 0)T , vL(k2) =√

2E(1, 0, 0, 0)T , (21.34)

while the polarization vectors for final photon and W− read

ε∗±µ(p1) = 1√2(0,− cos θ,±i, sin θ), ε∗±µ(p2) = 1√

2(0, cos θ,±i,− sin θ),

ε∗Lµ(p2) = 1mW

(p, EW sin θ, 0, EW cos θ). (21.35)

Then the amplitude is given by

iM(dLuR → γW−) =−ie2

√2sw

Ns

s−m2W

− ie2

3√

2sw

(−Nt

t+

2Nu

u

), (21.36)

21.4. Dependence of radiative corrections on the Higgs boson mass 185

where

Ns = vL(k2)[ε∗(p1) · ε∗(p2)(/p1

− /p2) + (p1 + 2p2) · ε∗(p1)/ε∗(p2)

− (2p1 + p2) · ε∗(p2)/ε∗(p1)]uL(k1), (21.37)

Nt = vL(k2)/ε∗(p2)(/k1 − /p1)/ε∗(p1)uL(k1), (21.38)

Nu = vL(k2)/ε∗(p1)(/k1 − /p2)/ε∗(p2)uL(k1). (21.39)

Now, using the physical conditions ε∗(pi) · pi = 0, /k1uL(k1) = 0 and vL(k2)/k2 = 0, we can

show that Ns = Nt −Nu. In fact,

Ns = vL(k2)[2ε∗(p1) · ε∗(p2)/p1

+ 2p2 · ε∗(p1)/ε∗(p2)− 2p1 · ε∗(p2)/ε∗(p1)]uL(k1),

Nt = vL(k2)[2k1 · ε1/ε2 + 2ε∗(p1) · ε∗(p2)/p1

− /ε∗(p1)/ε∗(p2)/p1

]uL(k1),

Nu = vL(k2)[− 2k2 · ε1/ε2 + 2p1 · ε∗(p2)/ε∗(p1)− /ε∗(p1)/ε∗(p2)/p1

]uL(k1).

Then Ns = Nt −Nu is manifest. Note further that s−m2W = −(t+ u), we have

iM(dLuR → γW−) =ie2

√2sw

(Nt −Nu

t+ u− Nt

3t+

2Nu

3u

)=

ie2

√2sw

(2t− u)

3(t+ u)

( Nt

t+Nu

u

)=

ie2

6√

2sw(1− 3 cos θ)

( Nt

t+Nu

u

). (21.40)

One can see clearly from this expression that all helicity amplitudes vanish at cos θ = 1/3.

(Note that the definition of scattering angle θ is different from the one in Peskin & Schroeder,

which, in our notation, is π − θ.) Then, by including all helicity combinations (6 in total),

we find the differential cross section, as a function of s and θ, to be

dσ

d cos θ=

πα2

32s2w

( 1− cos θ

sin θ

)2 x3 + 18x2 + 9x+ 24− (x3 − 14x2 + 9x− 8) cos 2θ

36(s−m2W )

, (21.41)

where x ≡ m2W/s.

21.4 Dependence of radiative corrections on the Higgs

boson mass

(a) We first analyze the radiative corrections to µ decay process at 1-loop level with the

Higgs boson in the loop. It is easy to see that if the internal Higgs boson line is attached to

one of the external fermions, the resulted vertex will contribute a factor of mf/v which can

be ignored. Therefore only the vacuum polarization diagrams are relevant, and they should

sum to a gauge invariant result.


(b) Now we compute the vacuum polarization amplitudes of W±, Z0 and photon with

Higgs contribution. We will only consider the pieces proportional to gµν , namely ΠWW (q2),

ΠZZ(q2), Πγγ(q2) and ΠZγ(q

2). It is easy to show that Πγγ(q2) and ΠZZ(q2) receive no

contribution from Higgs boson at 1-loop level, while ΠWW (q2) and ΠZZ(q2) can be found by

computing the following three diagrams:

h

Z0 Z0W+ W+

h

π0

h

Now we compute these three diagrams in turn for W+. The first diagram reads

(igmW )2gµν∫

d4k

(2π)4

i

k2 −m2h

−i

(q − k)2 −m2W

=− i

(4π)2g2m2

Wgµν

∫ 1

0

dxΓ(2− d

2)

∆2−d/2(m2W , q

2)

⇒− i

(4π)2g2m2

Wgµν

[E +

∫ 1

0

dx logM2

∆(m2W , q

2)

], (21.42)

where ∆(m2W , q

2) = xm2W + (1− x)m2

h− x(1− x)q2, E = 2/ε− γ+ log 4π− logM2, and M2

is the subtraction scale. The second one reads

(ig/2)2

∫d4k

(2π)4

i

k2 −m2h

−i

(q − k)2 −m2W

(2k − p)µ(2k − p)ν

⇒ g2

4gµν∫

d4k′

(2π)4

∫ 1

0

dx(4/d)k′2

(k′2 −∆(m2W , q

2))2

⇒ i

(4π)2

g2

4gµν∫ 1

0

dx 2∆(m2W , q

2)

[E + 1 + log

M2

∆(m2W , q

2)

], (21.43)

in which we have ignored terms proportional to qµqν . Then, the last diagram reads

1

2(ig2/2)gµν

∫d4k

(2π)4

i

k2 −m2h

= − g2

4gµν∫

d4k

(2π)4

1

k2 −m2h

(q − k)2 −m2W

(q − k)2 −m2W

=− g2

4gµν∫

d4k

(2π)4

∫ 1

0

dxk′2 + (1− x)2q2 −m2

W

(k′2 −∆(m2W , q

2))2

⇒− i

(4π)2

g2

4gµν∫ 1

0

dx

[(2∆(m2

W , q2)−m2

W + (1− x)2q2)E

+(

2∆(m2W , q

2)−m2W + (1− x)2q2

)log

M2

∆(m2W , q

2)+ ∆(m2

W , q2)

]. (21.44)

Thus we have, when the three diagrams above are taken into account only,

ΠWW (q2) =g2

4(4π)2

[−(

3m2W +

1

3q2)E

+

∫ 1

0

dx

(∆(m2

W , q2)−

[3m2

W + (1− x)2q2] logM2

∆(m2W , q

2)

)]. (21.45)

21.4. Dependence of radiative corrections on the Higgs boson mass 187

Now we extract Higgs mass contribution from this expression in the large Higgs limit, and

also fix the subtraction point at M2 = m2W . In this limit we may take ∆(m2

W , q2) ' xm2

h,

and log(M2/∆) ' − log(m2h/m

2W ). We also throw divergent terms with E, which should be

canceled out in the final expression of zeroth order natural relation after including completely

loop diagrams with W , Z, and would-be Goldstone boson internal lines. Then we have

ΠWW (q2) =g2

4(4π)2

[1

2m2h +

(3m2

W +1

3q2)

logm2h

m2W

]. (21.46)

Similarly, we have, for ΠZZ(q2),

ΠZZ(q2) =g2

4(4π)2 cos2 θw

[1

2m2h +

(3m2

Z +1

3q2)

logm2h

m2Z

]. (21.47)

(c) Now, we derive the zeroth order natural relation given in (21.134) of Peskin & Schroeder,

in the large Higgs mass limit. Note that Πγγ = ΠZγ = 0. Thus,

s2∗ − sin2 θ2

0 =sin2 θw cos2 θw

cos2 θw − sin2 θw

(ΠZZ(m2

Z)

m2Z

− ΠWW (0)

m2W

)=

α

48π

1 + 9 sin2 θwcos2 θw − sin2 θw

logm2h

m2W

, (21.48)

s2W − s2

∗ =− ΠWW (m2W )

m2Z

+m2W

m2Z

ΠZZ(m2Z)

m2Z

=5α

24πlog

m2h

m2W

. (21.49)


Final Project III

Decays of the Higgs Boson

In this final project, we calculate partial widths of various decay channels of the standard

model Higgs boson. Although a standard-model-Higgs-like boson has been found at the LHC

with mass around 125GeV, it is still instructive to treat the mass of the Higgs boson as a

free parameter in the following calculation.

The main decay modes of Higgs boson include h0 → ff with f the standard model

fermions, h0 → W+W−, h0 → Z0Z0, h0 → gg and h0 → γγ. The former three processes

appear at the tree level, while the leading order contributions to the latter two processes

are at one-loop level. We will work out the decay widths of these processes in the following.

In this problem we only consider the two-body final states. The calculation of decay

width needs the integral over the phase space of the two-body final states. By momentum

conservation and rotational symmetry, we can always parameterize the momenta of two final

particles in CM frame to be p1 = (E, 0, 0, p) and p2 = (E, 0, 0,−p), where E = 12mh by

energy conservation. Then the amplitude M will have no angular dependence. Then the

phase space integral reads ∫dΠ2 |M|2 =

1

4π

p

mh

|M|2. (21.50)

Then the decay width is given by

Γ =1

2mh

∫dΠ2 |M|2 =

1

8π

p

m2h

|M|2. (21.51)

In part (d) of this problem, we will also be dealing with the production of the Higgs

boson from two-gluon initial state, thus we also write down the formula here for the cross

section of the one-body final state from two identical initial particle. This time, the two

ingoing particles have momenta k1 = (E, 0, 0, k) and k2 = (E, 0, 0,−k), with E2 = k2 +m2i

and 2E = mf where mi and mf are masses of initial particles and final particle, respectively.

The final particle has momentum p = (mf , 0, 0, 0). Then, the cross section is given by

σ =1

2βs

∫d3p

(2π)3

1

2Ep|M|2(2π)4δ(4)(p− k1 − k2)

=1

4mfβs|M|2(2π)δ(2k −mf ) =

π

βm2f

|M|2δ(s−m2f ), (21.52)

189

190 Final Project III. Decays of the Higgs Boson

where β =√

1− (4mi/mf )2 is the magnitude of the velocity of the initial particle in the

center-of-mass frame.

(a) The easiest calculation of above processes is h0 → ff , where f represents all quarks

and charged leptons. The tree level contribution to this process involves a single Yukawa

vertex only. The corresponding amplitude is given by

iM(h0 → ff) = − imf

vu∗(p1)v(p2). (21.53)

Then it is straightforward to get the squared amplitude with final spins summed to be∑|M(h0 → ff)|2 =

m2f

v2tr[(/p1

+mf )(/p2−mf )

]=

2m2f

v2(m2

h − 4m2f ). (21.54)

In CM frame, the final states momenta can be taken to be p1 = (E, 0, 0, p) and p2 =

(E, 0, 0,−p), with E = 12mh and p2 = E2 −m2

f . Then the decay width is given by

Γ(h0 → ff) =1

8π

p

m2h

|M|2 =mhm

2f

8v2

(1−

4m2f

m2h

)3/2

. (21.55)

This expression can be expressed in terms of the fine structure constant α, the mass of W

boson mw and Weinberg angle sin θw, as

Γ(h0 → ff) =αmh

8 sin2 θw

m2f

m2W

(1−

4m2f

m2h

)3/2

. (21.56)

(b) Next we consider the decay of h0 to massive vector bosons W+W− and Z0Z0. The

amplitude for the process h0 → W+W− is given by

iM(h0 → W+W−) =igµνg2v

2ε∗µ(p1)ε∗ν(p2). (21.57)

Then the squared amplitude with final polarizations summed reads∑|M|2 =

g4v2

4

(gµν −

p1µp1ν

m2W

)(gµν − pµ2p

ν2

m2W

)=

πα

sin2 θw

m4h

m2W

(1− 4m2

W

m2h

+12m4

W

m4h

). (21.58)

Therefore the decay width is

Γ(h0 → W+W−) =1

8π

p1

m2h

|M|2 =αm3

h

16πm2W sin2 θw

(1− 4τ−1W + 12τ−2

W )(1− 4τ−1W )1/2, (21.59)

where we have defined τW ≡ (mh/mW )2 for brevity. For h0 → Z0Z0 process, it can be easily

checked that nothing gets changed in the calculation except that all mW should be replaced

with mZ , while an additional factor 1/2 is needed to account for the identical particles in

final state. Therefore we have

Γ(h0 → Z0Z0) =αm3

h

32πm2Z sin2 θw

(1− 4τ−1Z + 12τ−2

Z )(1− 4τ−1Z )1/2, (21.60)


where τZ ≡ (mh/mZ)2.

The calculation above considered “on-shell” decay only, while for realistic 125GeV boson,

the off-shell decay turns out to be very important. That is, although h0 → W+W− and

h0 → Z0Z0 are kinetically forbidden when mh = 125GeV according to above results, the

produced W or Z pair can subsequently decay into lighter fermions, and the process like

h0 → W ∗(ff)W contributes considerable amount of Higgs decay, where W ∗(ff) means an

off-shell W decaying to a pair of fermions. More details can be found in [8]

(c) Now we come to the process h0 → gg. The leading order contribution comes from

diagrams with one quark loop.

The amplitude reads

iM(h0 → gg) =− imq

v(igs)

2ε∗µ(p1)ε∗ν(p2) tr (tatb)

×∫

ddq

(2π)d

(−1) tr

[γµ

i

/q −mq

γνi

/q + /p2−mq

i

/q − /p1−mq

]+ (−1) tr

[γν

i

/q −mq

γµi

/q + /p1−mq

i

/q − /p2−mq

](21.61)

The first trace in the integrand can be simplified through standard procedure,

tr

[γµ

i

/q −mq

γνi

/q + /p2−mq

i

/q − /p1−mq

]=−i tr

[(/q +mq)(/q + /p2

−mq)(/q − /p1−mq)

](q2 −m2

q)[(q + p2)2 −m2

q

][(q − p1)2 −m2

q

]= −2i

∫ 1

0

dx

∫ 1−x

0

dyNµν

(q′2 −∆)3, (21.62)

where

q′µ = qµ − xp1µ + yp2µ, (21.63)

∆ = m2q − x(1− x)p2

1 − y(1− y)p22 − 2xyp1 · p2 = m2

q − xym2h,

Nµν = 4mq

(pν1p

ν2 − p

µ1p

ν2 + 2pν2q

µ − 2pµ1qν + 4qµqν + (m2

q − p1 · p2 − q2)ηµν)

(21.64)

Then we can reexpress Nµν in terms of q′, p1 and p2 and drop off all terms linear in q which

integrates to zero. It is most easy to work with definite helicity states for final gluons. Then

the result gets simplified if we dot Nµν with polarization vectors as Nµνε∗µ(p1)ε∗ν(p2). Note


that ε∗(pi)·pj = 0 with i, j = 1, 2. Note also the on-shell condition p21 = p2

2 = 0, p1·p2 = 12m2h.

then

Nµνε∗µ(p1)ε∗ν(p2) = 4mq

[m2q +

(xy − 1

2

)m2h +

(4d− 1)q′2]ε∗(p1) · ε∗(p2). (21.65)

The same calculation shows that the second trace in the integrand of (21.61) gives identical

result with the first trace. To check the gauge invariance of this result, one can simply

replace ε∗µ(p1) with p1µ in the expression above, then it is straightforward to find that

Nµνp1µε∗ν(p2) = 0. Similarly, it can also be checked that Nµνε∗µ(p1)p2ν = 0.

Then the amplitude (21.61) now reads

iM(h0 → gg) = − 2g2smq

vδab∫ 1

0

dx

∫ 1−x

0

dy

∫ddq′

(2π)dNµνε∗µ(p1)ε∗ν(p2)

(q′2 −∆)3, (21.66)

where the relation tr (tatb) = 12δab in fundamental representation is also used. The momen-

tum integration is finite as d→ 4 under dimensional regularization, and can now be carried

out directly to be

iM(h0 → gg) =−2ig2

sm2q

(4π)2vδabε∗(p1) · ε∗(p2)

∫ 1

0

dx

∫ 1−x

0

dy(1− 4xy)m2

h

m2q − xym2

h

=− iαsm2h

6πvδabε∗(p1) · ε∗(p2)If (τq), (21.67)

where τq ≡ (mh/mq)2, and

If (τq) ≡ 3

∫ 1

0

dx

∫ 1−x

0

dy1− 4xy

1− xyτq

Note that the inner product between two polarization vectors is nonzero only for ε∗+ · ε∗−and ε∗− · ε∗+. Therefore the squared amplitude with final states polarizations, color indices

summed (δabδab = 8) is,

|M(h0 → gg)|2 = |M+−(h0 → gg)|2 + |M−+(h0 → gg)|2 =4α2

sm4h

9π2v2|If (τq)|2, (21.68)

and the decay width is

Γ(h0 → gg) =( αmh

8 sin2 θw

)· m

2h

m2w

· α2s

9π2· |If (τq)|2, (21.69)

where an additional factor 1/2 should be included in (21.51) when calculating Γ(h0 → gg)

because the two gluons in final states are identical particles. This result is easily generalized

for Nq copies of quarks to be

Γ(h0 → gg) =( αmh

8 sin2 θw

)· m

2h

m2W

· α2s

9π2·∣∣∣∑

q

If (τq)∣∣∣2, (21.70)


(d) Now we calculate the cross section for the Higgs production via gluon fusion at the

leading order. The amplitude is simply given by the result in (c), namely (21.67). When

we take the square of this amplitude, an additional factor ( 18· 1

2)2 should be included, to

average over helicities and color indices of initial gluons. Then, comparing (21.52) with

(21.51), we find that

σ(gg → h0) =π2

8mh

δ(s−m2h)Γ(h0 → gg), (21.71)

where the hatted variable s is the parton level center-of-mass energy. We note again that

the correct formula is obtained by including an factor of ( 18· 1

2)2 in σ(gg → h0) to average

over the initial degrees of freedom of two gluons, and an factor of 1/2 in Γ(h0 → gg) to

count the identical particles in the final state. Then, from (21.70), it is straightforward to

find

σ(gg → h0) =αα2

s

576 sin2 θw· m

2h

m2W

∣∣∣∑q

If (τq)∣∣∣2δ(s−m2

h). (21.72)

Then the proton-level cross section of Higgs boson production via gluon-gluon fusion is given

by

σGGF(p(P1)p(P2)→ h0)

=

∫ 1

0

dx1

∫ 1

0

dx2 fg(x1)fg(x2)σ(g(x1P1)g(x2P2)→ h0

)=

∫dM2dY

∣∣∣∣ ∂(x1, x2)

∂(M2, Y )

∣∣∣∣fg(x1)fg(x2)σ(g(x1P1)g(x2P2)→ h0

)=

∫dM2dY

1

M2x1fg(x1)x2fg(x2)σ

(g(x1P1)g(x2P2)→ h0

), (21.73)

where M2 = x1x2s is the center-of-mass energy of two initial gluons, while s is the center-of-

mass energy of two initial protons, and Y , given by expY =√x1/x2, is the rapidity of the

produced Higgs boson relative to the center-of-mass frame of the proton system. (Note that

in our case M2 = m2h.) The relations between M2, Y and the momentum fractions x1, x2 can

be inverted to give x1 = (M/√s)eY and x2 = (M/

√s)e−Y . Furthermore, fg is the parton

distribution function of the gluon in a proton, which we will take to be fg = 8(1− x)7/x in

the following calculations. Then the cross section can be evaluated to be

σGGF(p(P1)p(P2)→ h0)

=αα2

s

9 sin2 θw· 1

m2W

∣∣∣∑q

If (τq)∣∣∣2 ∫ Y0

−Y0dY(

1− mh√seY)7(

1− mh√se−Y

)7

, (21.74)

where Y0, given by coshY0 =√s/2mh is the largest possible rapidity of a produced Higgs

boson. We plot this cross section as a function of the center-of-mass energy√s of the pp

pair, with the Higgs boson’s mass taken to be mh = 30GeV and mh = 125GeV, respectively,

in Figure


1 2 5 10 20 501

2

5

10

20

50

s GeV

Cro

ssS

ecti

onp

b mh=30 GeV

mh=125 GeV

(e) Next we consider the process h0 → 2γ. The contribution to this decay channel at the

leading (1-loop) level is from two types of diagrams, one with a fermion loop and the other

with a W boson (and related would-be Goldstone boson) loop. The former contribution is

easy to find by virtue of the result in (c) for h0 → gg. The calculation here is in fully parallel,

except that we should include the factor for the electric charges of internal fermions Qf , take

away the color factor tr (tatb), change the strong coupling gs by the electromagnetic coupling

e, and sum over all charged fermions. Note that the color factor enters the expression of

the decay width as | tr (tatb)|2 = 12δab 1

2δab = 2, then it is straightforward to write down the

fermion contribution to the h0 → 2γ to be

iM(h0 → 2γ)f =( αmh

8 sin2 θw

)· m

2h

m2w

· α2s

18π2·∣∣∣∑

f

Q2fNc(f)If (τf )

∣∣∣2, (21.75)

where Nc(f) is the color factor, equal to 3 for quarks and 1 for charged leptons.

(f) Now we come to the W -loop contributions to h0 → 2γ. In Feynman-’t Hooft gauge, we

should also include the corresponding Goldstone loop diagrams. Then there are 13 diagrams

in total. We compute them as follows,

(a) (b)


iM(a) =1

2

igρσg2v

2(−ie2)(2ηµνηρσ − ηµρηνσ − ηµσηνρ)ε∗µ(p1)ε∗ν(p2)

×∫

ddq

(2π)dDW (q)DW (k − q)

=− 2i

(4π)d/2e2m2

W

vε∗(p1) · ε∗(p2)(d− 1)Γ(2− d

2)

×∫ 1

0

dx

[m2W − x(1− x)m2

h]2−d/2 , (21.76)

iM(b) =1

2(−2iλv)(2ie2)ε∗(p1) · ε∗(p2)

∫ddq

(2π)dDs(q)Ds(k − q)

=− i

(4π)d/2e2m2

h

vε∗(p1) · ε∗(p2)Γ(2− d

2)

×∫ 1

0

dx

[m2W − x(1− x)m2

h]2−d/2 . (21.77)

(c) (d)

iM(c) = iM(d) =ig2 sin θw

2· ig2v sin θw

2ε∗(p1) · ε∗(p2)

∫ddq

(2π)dDs(q)DW (p2 − q)

=− i

(4π)d/2e2m2

W

vε∗(p1) · ε∗(p2)Γ(2− d

2)

1

(m2W )2−d/2 . (21.78)

(e) (f) (g)

(h) (i) (j)

(k) (l) (m)


iM(e) =ig2v

2(−ie)2ηρσε

∗µ(p1)ε∗ν(p2)

∫ddq

(2π)dDW (q)DW (q − p1)DW (q + p2)

×[ηρλ(2q − p1)µ + ηµρ(2p1 − q)λ − ηλµ(p1 + q)ρ

]×[ησλ(2q + p2)ν − ηνλ(q − p2)σ − ησν(2p2 + q)λ

]=

i

(4π)d/2e2m2

W

vε∗(p1) · ε∗(p2)

[ ∫dxdy

(5− x− y + 4xy)m2h

m2W − xym2

h

+ 6(d− 1)Γ(2− d2

)

∫dxdy

(m2W − xym2

h)2−d/2

], (21.79)

iM(f) = (−2iλν)(−ie)2ε∗µ(p1)ε∗ν(p2)

∫ddq

(2π)d(2q − p1)µ(2q + p2)ν

×Ds(q)Ds(q − p1)Ds(q + p2)

=i

(4π)d/2e2m2

h

vε∗(p1) · ε∗(p2)Γ(2− d

2)

∫2dxdy

(m2W − xym2

h)2−d/2 , (21.80)

iM(g) =(− im2

W

v

)(ie)2ε∗µ(p1)ε∗ν(p2)

∫ddq

(2π)d(−1)(q − p1)µqν

×Ds(q)Ds(q − p1)Ds(q + p2)

=− i

(4π)d/2e2m2

W

vε∗(p1) · ε∗(p2)Γ(2− d

2)

∫dxdy

(m2W − xym2

h)2−d/2 , (21.81)

iM(h) = iM(i) =ig

2

igµλg2v sin θw2

(−ie)ε∗µ(p1)ε∗ν(p2)

∫ddq

(2π)d(q − p1 − k)σ

×[ησλ(2q + p2)ν − ηνλ(q − p2)σ − ηνσ(2p2 + q)λ

]×DW (q)Ds(q − p1)DW (q + p2)

=i

(4π)d/2e2m2

W

vε∗(p1) · ε∗(p2)

[ ∫dxdy

(1− x)(1 + y)m2h

m2W − xym2

h

− 1

2(d− 1)Γ(2− d

2)

∫dxdy

(m2W − xym2

h)2−d/2

], (21.82)

iM(j) =ig2v

2

( ig2v sin θw2

)2

ε∗(p1) · ε∗(p2)

×∫

ddq

(2π)dDs(q)DW (q − p1)DW (q + p2)

=i

(4π)d/2e2m2

W

vε∗(p1) · ε∗(p2)

∫dxdy

2m2W

m2W − xym2

h

, (21.83)

iM(k) = iM(l) =ig

2

ig2v sin θw2

(−ie)ε∗µ(p1)ε∗ν(p2)

×∫

ddq

(2π)d(p1 + 2p2 + q)µ(2q + p2)νDs(q)DW (q − p1)Ds(q + p2)

=i

(4π)d/2e2m2

W

vε∗(p1) · ε∗(p2)Γ(2− d

2)

∫dxdy

(m2W − xym2

h)2−d/2 , (21.84)


iM(m) = (−2iλν)( ig2v sin θw

2

)2

ε∗(p1) · ε∗(p2)

×∫

ddq

(2π)dDW (q)Ds(q − p1)Ds(q + p2)

=i

(4π)d/2e2m2

h

vε∗(p1) · ε∗(p2)

∫dxdy

m2W

m2W − xym2

h

. (21.85)

The results can be summarized as,

iM(X) =i

(4π)d/2e2m2

W

vε∗(p1) · ε∗(p2)

[A · Γ(2− d

2) +B

], (X = a, b, · · · ,m) (21.86)

with the coefficients A and B for each diagram listed in Table.

Diagrams A B

(a) −2(d− 1)J1 0

(c)+(d) −2(m2W )d/2−2 0

(e) 6(d− 1)J2 J3

(g) −J2 0

(h)+(i) −(d− 1)J2 2J4

(j) 0 2(mW/mh)2J5

(k)+(l) 2J2 0

(m) 0 J5

(b) −(mh/mW )2J1 0

(f) 2(mh/mW )2J2 0

where

J1 =

∫ 1

0

dx1

[m2W − x(1− x)m2

h]2−d/2

= 1− ε

2

∫ 1

0

dx log(m2W − x(1− x)m2

h

)+O(ε2), (21.87)

J2 =

∫ 1

0

dx

∫ 1−x

0

dy1

(m2W − xym2

h)2−d/2

=1

2− ε

2

∫ 1

0

dx

∫ 1−x

0

dy log(m2W − xym2

h

)+O(ε2), (21.88)

J3 =

∫ 1

0

dx

∫ 1−x

0

dy(5− x− y + 4xy)m2

h

m2W − xym2

h

, (21.89)

J4 =

∫ 1

0

dx

∫ 1−x

0

dy(1− x)(1 + y)m2

h

m2W − xym2

h

, (21.90)

J5 =

∫ 1

0

dx

∫ 1−x

0

dym2h

m2W − xym2

h

. (21.91)

To see that the divergences of all diagrams cancel among themselves, it just needs to show

that sum of all A-coefficients is of order ε. This is straightforward by noting that J1 =

1 +O(ε) and J2 = 1/2 +O(ε).


Before reaching the complete result, let us first find out the W -loop contribution in the

limit m2h m2

W , although it seems unlikely to be true within our current knowledge. To

find the amplitude in this limit, we expand the five integrals J1, · · · , J5 in terms of mh/mW ,

J1 ' 1− ε

2logm2

W +ε

12

m2h

m2W

, J2 '1

2− ε

4logm2

W +ε

48

m2h

m2W

,

J3 '7

3

m2h

m2W

, J4 '11

24

m2h

m2W

, J5 '1

2

m2h

m2W

+1

24

( m2h

m2W

)2

.

Then the amplitude can be recast into

iM =ie2m2

W

(4π)2vε∗(p1) · ε∗(p2)

[C( 2

ε− γ + log 4π

)+D · logm2

W + E + F · m2h

m2W

](21.92)

Diagrams C D E F

(a) −6 3 4 −1

(c)+(d) −2 1 0 0

(e) 9 −9/2 −6 37/12

(g) −1/2 1/4 0 −1/24

(h)+(i) −3/2 3/4 1 19/24

(j) 0 0 1 1/12

(k)+(l) 1 −1/2 0 1/12

(m) 0 0 0 1/2

(b) −(mh/mW )2 (mh/mW )2/2 0 0

(f) (mh/mW )2 −(mh/mW )2/2 0 0

sum 0 0 0 7/2

Therefore, the amplitude in the limit m2h m2

W is given by

iM(h0 → 2γ)W = 2 · 7

2

iαm2h

4πvε∗(p1) · ε∗(p2), (21.93)

where the factor 2 counts the identical contributions from the diagrams with two final

photons changed. Now we sum up the fermion-loop contribution found in (e) and the result

here to get the h0 → 2γ amplitude in the light Higgs limit,

iM = − iαm2h

3πv

[∑f

Q2fNc(f)− 21

4

]ε∗(p1) · ε∗(p2). (21.94)

Then the corresponding partial width is given by

Γ(h0 → 2γ) =( αmh

8 sin2 θw

)· m

2h

m2w

· α2

18π2·∣∣∣∣∑

f

Q2fNc(f)− 21

4

∣∣∣∣2, (21.95)

Now we retain mh as a free variable. Then the various diagrams sum into the following

full expression for the W -loop contribution to h0 → 2γ,

iM(h0 → 2γ)W =iαm2

h

2πvε∗(p1) · ε∗(p2)IW (τW ), (21.96)


where the factor IW (τW ), as a function of τW ≡ (mh/mW )2, is given by

IW (τW ) =1

τW

[6I1(τW )− 8I2(τW ) + τW

(I1(τW )− I2(τW )

)+ I3(τW )

], (21.97)

where

I1(τW ) ≡∫ 1

0

dx log[1− x(1− x)τW

], (21.98)

I2(τW ) ≡ 2

∫ 1

0

dx

∫ 1−x

0

dy log(1− xyτW

), (21.99)

I3(τW ) ≡∫ 1

0

dx

∫ 1−x

0

dy(8− 3x+ y + 4xy)τW

1− xyτW. (21.100)

Then the full expression for the partial width of h0 → 2γ at one-loop is

Γ(h0 → 2γ) =( αmh

8 sin2 θw

)· m

2h

m2w

· α2

18π2·∣∣∣∣∑

f

Q2fNc(f)If (τf )− IW (τW )

∣∣∣∣2, (21.101)

(h) Collecting all results above (expect the γγ channel, which is quite small∗), we plot

the total width and decay branching fractions of the Higgs boson in Figures 21.6 and 21.7,

respectively.

100 200 300 400 50010-4

0.001

0.01

0.1

1

10

mhGeV

To

tal

Wid

thG

eV

Figure 21.6: The total width of the Higgs boson as a function of its mass.

∗– but very important!


100 200 300 400 5000.001

0.005

0.010

0.050

0.100

0.500

1.000

mh HGeVL

Bra

nch

ing

Rat

ios

W+W-

Z0Z0

tt

gg

bb

cc

Τ+Τ-

Figure 21.7: The Higgs decay branching fractions of tt, bb, cc, τ+τ−, WW , ZZ and gg

channels, as functions of Higgs mass.

Bibliography

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[3] K. A. Olive et al. (Particle Data Group), Chin. Phys. C, 38 (2014) 090001.

[4] A. Zee, “Quantum Field Theory in a Nutshell”, 2nd edition, Princeton University Press,

2010.

[5] S. Coleman, “Aspects of Symmetry”, Cambridge University Press, 1985.

[6] S. R. Coleman and E. J. Weinberg, “Radiative Corrections as the Origin of Spontaneous

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201

Science

A complete solution to problems in an introduction to quantum field theory by peskin and schroeder